What is the general solution for cos^6(x)=-cos^2(x) ?

The general solution for cos^6(x)=-cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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Popular Trigonometry >

cos^6(x)=-cos^2(x)

Solution

cos^{6} (x) = - cos^{2} (x)

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos^{6} (x) = - cos^{2} (x)

Solve by substitution

cos^{6} (x) = - cos^{2} (x)

Let:

cos (x) = u

u^{6} = - u^{2}

u^{6} = - u^{2} : u = 0, u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i, u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

u^{6} = - u^{2}

Move

u^{2}

to the left side

u^{6} = - u^{2}

Add

u^{2}

to both sides

u^{6} + u^{2} = - u^{2} + u^{2}

Simplify

u^{6} + u^{2} = 0

u^{6} + u^{2} = 0

Rewrite the equation with

a = u^{2}

and

a^{3} = u^{6}

a^{3} + a = 0

Solve

a^{3} + a = 0 : a = 0, a = i, a = - i

a^{3} + a = 0

Factor

a^{3} + a : a (a^{2} + 1)

a^{3} + a

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

a^{3} = a^{2} a

= a^{2} a + a

Factor out common term

a

= a (a^{2} + 1)

a (a^{2} + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

a = 0 or a^{2} + 1 = 0

Solve

a^{2} + 1 = 0 : a = i, a = - i

a^{2} + 1 = 0

Move

1

to the right side

a^{2} + 1 = 0

Subtract

1

from both sides

a^{2} + 1 - 1 = 0 - 1

Simplify

a^{2} = - 1

a^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

a = - 1, a = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

a = i, a = - i

The solutions are

a = 0, a = i, a = - i

a = 0, a = i, a = - i

Substitute back

a = u^{2},

solve for

u

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

Solve

u^{2} = i : u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i

u^{2} = i

Substitute

u = a + bi

(a + bi)^{2} = i

Expand

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} b^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= (- 1) b^{2}

Refine

= - b^{2}

= a^{2} + 2 iab - b^{2}

Rewrite

a^{2} + 2 iab - b^{2}

in standard complex form:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

Group the real part and the imaginary part of the complex number

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = i

Rewrite

i

in standard complex form:

0 + i

(a^{2} - b^{2}) + 2 iab = 0 + i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[a^{2} - b^{2} = 0 2 ab = 1]

[a^{2} - b^{2} = 0 2 ab = 1] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

[a^{2} - b^{2} = 0 2 ab = 1]

Isolate

a

for

2 ab = 1 : a = \frac{1}{2 b}

2 ab = 1

Divide both sides by

2 b

2 ab = 1

Divide both sides by

2 b

\frac{2 ab}{2 b} = \frac{1}{2 b}

Simplify

a = \frac{1}{2 b}

a = \frac{1}{2 b}

Plug the solutions

a = \frac{1}{2 b}

into

a^{2} - b^{2} = 0

For

a^{2} - b^{2} = 0

, subsitute

a

with

\frac{1}{2 b} : b = \frac{1}{2}, b = - \frac{1}{2}

For

a^{2} - b^{2} = 0

, subsitute

a

with

\frac{1}{2 b}

(\frac{1}{2 b})^{2} - b^{2} = 0

Solve

(\frac{1}{2 b})^{2} - b^{2} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

(\frac{1}{2 b})^{2} - b^{2} = 0

Simplify

(\frac{1}{2 b})^{2} : \frac{1}{4 b ^{2}}

(\frac{1}{2 b})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 2 b ) ^{2}}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2 b)^{2} = 2^{2} b^{2}

= \frac{1 ^{2}}{2 ^{2} b ^{2}}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{2 ^{2} b ^{2}}

2^{2} = 4

= \frac{1}{4 b ^{2}}

\frac{1}{4 b ^{2}} - b^{2} = 0

Multiply both sides by

4 b^{2}

\frac{1}{4 b ^{2}} - b^{2} = 0

Multiply both sides by

4 b^{2}

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

Simplify

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

Simplify

\frac{1}{4 b ^{2}} \cdot 4 b^{2} : 1

\frac{1}{4 b ^{2}} \cdot 4 b^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4 b ^{2}}{4 b ^{2}}

Cancel the common factor:

4

= \frac{1 \cdot b ^{2}}{b ^{2}}

Cancel the common factor:

b^{2}

= 1

Simplify

- b^{2} \cdot 4 b^{2} : - 4 b^{4}

- b^{2} \cdot 4 b^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

b^{2} b^{2} = b^{2 + 2}

= - 4 b^{2 + 2}

Add the numbers:

2 + 2 = 4

= - 4 b^{4}

Simplify

0 \cdot 4 b^{2} : 0

0 \cdot 4 b^{2}

Apply rule

0 \cdot a = 0

= 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

Solve

1 - 4 b^{4} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

1 - 4 b^{4} = 0

Move

1

to the right side

1 - 4 b^{4} = 0

Subtract

1

from both sides

1 - 4 b^{4} - 1 = 0 - 1

Simplify

- 4 b^{4} = - 1

- 4 b^{4} = - 1

Divide both sides by

- 4

- 4 b^{4} = - 1

Divide both sides by

- 4

\frac{- 4 b ^{4}}{- 4} = \frac{- 1}{- 4}

Simplify

b^{4} = \frac{1}{4}

b^{4} = \frac{1}{4}

For

x^{n} = f (a)

, n is even, the solutions are

Apply radical rule:

Factor the number:

4 = 2^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

Apply radical rule:

assuming

a \geq 0

= 2

= \frac{1}{2}

Apply radical rule:

Factor the number:

4 = 2^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

Apply radical rule:

assuming

a \geq 0

= 2

= - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

b = 0

Take the denominator(s) of

(\frac{1}{2 b})^{2} - b^{2}

and compare to zero

Solve

2 b = 0 : b = 0

2 b = 0

Divide both sides by

2

2 b = 0

Divide both sides by

2

\frac{2 b}{2} = \frac{0}{2}

Simplify

b = 0

b = 0

The following points are undefined

b = 0

Combine undefined points with solutions:

b = \frac{1}{2}, b = - \frac{1}{2}

Plug the solutions

b = \frac{1}{2}, b = - \frac{1}{2}

into

2 ab = 1

For

2 ab = 1

, subsitute

b

with

\frac{1}{2} : a = \frac{1}{2}

For

2 ab = 1

, subsitute

b

with

\frac{1}{2}

2 a \frac{1}{2} = 1

Solve

2 a \frac{1}{2} = 1 : a = \frac{1}{2}

2 a \frac{1}{2} = 1

Multiply both sides by

2

2 a \frac{1}{2} = 1

Multiply both sides by

2

2 a \frac{1}{2} 2 = 1 \cdot 2

Simplify

2 a \frac{1}{2} 2 = 1 \cdot 2

Simplify

2 a \frac{1}{2} 2 : 2 a

2 a \frac{1}{2} 2

Convert

2

to fraction

: \frac{2}{1}

2

Convert element to fraction:

2 = \frac{2}{1}

= \frac{2}{1}

= 2 a \frac{1}{2} \cdot \frac{2}{1}

Cross-cancel common factor:

2

= 2 a \cdot 1

Apply rule:

a \cdot 1 = a

= 2 a

Simplify

1 \cdot 2 : 2

1 \cdot 2

Apply rule:

1 \cdot a = a

= 2

2 a = 2

2 a = 2

2 a = 2

Divide both sides by

2

2 a = 2

Divide both sides by

2

\frac{2 a}{2} = \frac{2}{2}

Simplify

\frac{2 a}{2} = \frac{2}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Cancel the common factor:

2

= a

Simplify

\frac{2}{2} : \frac{1}{2}

\frac{2}{2}

Apply radical rule:

a = a a

2 = 22

= \frac{2}{2 2}

Cancel the common factor:

2

= \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

For

2 ab = 1

, subsitute

b

with

- \frac{1}{2} : a = - \frac{1}{2}

For

2 ab = 1

, subsitute

b

with

- \frac{1}{2}

2 a (- \frac{1}{2}) = 1

Solve

2 a (- \frac{1}{2}) = 1 : a = - \frac{1}{2}

2 a (- \frac{1}{2}) = 1

Divide both sides by

2 (- \frac{1}{2})

2 a (- \frac{1}{2}) = 1

Divide both sides by

2 (- \frac{1}{2})

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{1}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{1}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : a

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

Apply rule:

a (- b) = - ab

2 a (- \frac{1}{2}) = - 2 a \frac{1}{2}

= \frac{- 2 a \frac{1}{2}}{2 ( - \frac{1}{2} )}

Apply rule:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

= \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

= \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

Cancel the common factor:

- 2

= \frac{a \frac{1}{2}}{\frac{1}{2}}

Cancel the common factor:

\frac{1}{2}

= a

Simplify

\frac{1}{2 ( - \frac{1}{2} )} : - \frac{1}{2}

\frac{1}{2 ( - \frac{1}{2} )}

Apply rule:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

= \frac{1}{- 2 \cdot \frac{1}{2}}

- 2 \cdot \frac{1}{2} = - 2

- 2 \cdot \frac{1}{2}

Convert

2

to fraction

: \frac{2}{1}

2

Convert element to fraction:

2 = \frac{2}{1}

= \frac{2}{1}

= - \frac{2}{1} \cdot \frac{1}{2}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

\frac{2}{1} \cdot \frac{1}{2} = \frac{2 \cdot 1}{1 \cdot 2}

= - \frac{2 \cdot 1}{1 \cdot 2}

\frac{2 \cdot 1}{1 \cdot 2} = 2

\frac{2 \cdot 1}{1 \cdot 2}

\frac{2 \cdot 1}{1 \cdot 2} = \frac{2}{2}

\frac{2 \cdot 1}{1 \cdot 2}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{1 \cdot 2}

Apply rule:

1 \cdot a = a

1 \cdot 2 = 2

= \frac{2}{2}

= \frac{2}{2}

Apply radical rule:

a = a a

2 = 22

= \frac{2 2}{2}

Cancel the common factor:

2

= 2

= - 2

= \frac{1}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

Verify solutions by plugging them into the original equations

Check the solutions by plugging them into

a^{2} - b^{2} = 0

Remove the ones that don't agree with the equation.

Check the solution

a = - \frac{1}{2}, b = - \frac{1}{2} :

True

a^{2} - b^{2} = 0

Plug in

a = - \frac{1}{2}, b = - \frac{1}{2}

(- \frac{1}{2})^{2} - (- \frac{1}{2})^{2} = 0

Refine

0 = 0

True

Check the solution

a = \frac{1}{2}, b = \frac{1}{2} :

True

a^{2} - b^{2} = 0

Plug in

a = \frac{1}{2}, b = \frac{1}{2}

(\frac{1}{2})^{2} - (\frac{1}{2})^{2} = 0

Refine

0 = 0

True

Check the solutions by plugging them into

2 ab = 1

Remove the ones that don't agree with the equation.

Check the solution

a = - \frac{1}{2}, b = - \frac{1}{2} :

True

2 ab = 1

Plug in

a = - \frac{1}{2}, b = - \frac{1}{2}

2 (- \frac{1}{2}) (- \frac{1}{2}) = 1

Refine

1 = 1

True

Check the solution

a = \frac{1}{2}, b = \frac{1}{2} :

True

2 ab = 1

Plug in

a = \frac{1}{2}, b = \frac{1}{2}

2 \cdot \frac{1}{2} \cdot \frac{1}{2} = 1

Refine

1 = 1

True

Therefore, the final solutions for

a^{2} - b^{2} = 0, 2 ab = 1

are

(a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

Substitute back

u = a + bi

u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i

Solve

u^{2} = - i : u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

u^{2} = - i

Substitute

u = a + bi

(a + bi)^{2} = - i

Expand

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} b^{2}

i^{2} = - 1

i^{2}

Apply imaginary number rule:

i^{2} = - 1

= - 1

= (- 1) b^{2}

Refine

= - b^{2}

= a^{2} + 2 iab - b^{2}

Rewrite

a^{2} + 2 iab - b^{2}

in standard complex form:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

Group the real part and the imaginary part of the complex number

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = - i

Rewrite

- i

in standard complex form:

0 - i

(a^{2} - b^{2}) + 2 iab = 0 - i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[a^{2} - b^{2} = 0 2 ab = - 1]

[a^{2} - b^{2} = 0 2 ab = - 1] : (a = - \frac{1}{2}, a = \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

[a^{2} - b^{2} = 0 2 ab = - 1]

Isolate

a

for

2 ab = - 1 : a = - \frac{1}{2 b}

2 ab = - 1

Divide both sides by

2 b

2 ab = - 1

Divide both sides by

2 b

\frac{2 ab}{2 b} = \frac{- 1}{2 b}

Simplify

a = - \frac{1}{2 b}

a = - \frac{1}{2 b}

Plug the solutions

a = - \frac{1}{2 b}

into

a^{2} - b^{2} = 0

For

a^{2} - b^{2} = 0

, subsitute

a

with

- \frac{1}{2 b} : b = \frac{1}{2}, b = - \frac{1}{2}

For

a^{2} - b^{2} = 0

, subsitute

a

with

- \frac{1}{2 b}

(- \frac{1}{2 b})^{2} - b^{2} = 0

Solve

(- \frac{1}{2 b})^{2} - b^{2} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

(- \frac{1}{2 b})^{2} - b^{2} = 0

Simplify

(- \frac{1}{2 b})^{2} : \frac{1}{4 b ^{2}}

(- \frac{1}{2 b})^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- \frac{1}{2 b})^{2} = (\frac{1}{2 b})^{2}

= (\frac{1}{2 b})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 2 b ) ^{2}}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2 b)^{2} = 2^{2} b^{2}

= \frac{1 ^{2}}{2 ^{2} b ^{2}}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{2 ^{2} b ^{2}}

2^{2} = 4

= \frac{1}{4 b ^{2}}

\frac{1}{4 b ^{2}} - b^{2} = 0

Multiply both sides by

4 b^{2}

\frac{1}{4 b ^{2}} - b^{2} = 0

Multiply both sides by

4 b^{2}

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

Simplify

\frac{1}{4 b ^{2}} \cdot 4 b^{2} - b^{2} \cdot 4 b^{2} = 0 \cdot 4 b^{2}

Simplify

\frac{1}{4 b ^{2}} \cdot 4 b^{2} : 1

\frac{1}{4 b ^{2}} \cdot 4 b^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4 b ^{2}}{4 b ^{2}}

Cancel the common factor:

4

= \frac{1 \cdot b ^{2}}{b ^{2}}

Cancel the common factor:

b^{2}

= 1

Simplify

- b^{2} \cdot 4 b^{2} : - 4 b^{4}

- b^{2} \cdot 4 b^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

b^{2} b^{2} = b^{2 + 2}

= - 4 b^{2 + 2}

Add the numbers:

2 + 2 = 4

= - 4 b^{4}

Simplify

0 \cdot 4 b^{2} : 0

0 \cdot 4 b^{2}

Apply rule

0 \cdot a = 0

= 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

1 - 4 b^{4} = 0

Solve

1 - 4 b^{4} = 0 : b = \frac{1}{2}, b = - \frac{1}{2}

1 - 4 b^{4} = 0

Move

1

to the right side

1 - 4 b^{4} = 0

Subtract

1

from both sides

1 - 4 b^{4} - 1 = 0 - 1

Simplify

- 4 b^{4} = - 1

- 4 b^{4} = - 1

Divide both sides by

- 4

- 4 b^{4} = - 1

Divide both sides by

- 4

\frac{- 4 b ^{4}}{- 4} = \frac{- 1}{- 4}

Simplify

b^{4} = \frac{1}{4}

b^{4} = \frac{1}{4}

For

x^{n} = f (a)

, n is even, the solutions are

Apply radical rule:

Factor the number:

4 = 2^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

Apply radical rule:

assuming

a \geq 0

= 2

= \frac{1}{2}

Apply radical rule:

Factor the number:

4 = 2^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

2^{2} = 2^{0.5 \cdot 4} = (2)^{4}

Apply radical rule:

assuming

a \geq 0

= 2

= - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

b = 0

Take the denominator(s) of

(- \frac{1}{2 b})^{2} - b^{2}

and compare to zero

Solve

2 b = 0 : b = 0

2 b = 0

Divide both sides by

2

2 b = 0

Divide both sides by

2

\frac{2 b}{2} = \frac{0}{2}

Simplify

b = 0

b = 0

The following points are undefined

b = 0

Combine undefined points with solutions:

b = \frac{1}{2}, b = - \frac{1}{2}

Plug the solutions

b = \frac{1}{2}, b = - \frac{1}{2}

into

2 ab = - 1

For

2 ab = - 1

, subsitute

b

with

\frac{1}{2} : a = - \frac{1}{2}

For

2 ab = - 1

, subsitute

b

with

\frac{1}{2}

2 a \frac{1}{2} = - 1

Solve

2 a \frac{1}{2} = - 1 : a = - \frac{1}{2}

2 a \frac{1}{2} = - 1

Multiply both sides by

2

2 a \frac{1}{2} = - 1

Multiply both sides by

2

2 a \frac{1}{2} 2 = (- 1) 2

Simplify

2 a \frac{1}{2} 2 = (- 1) 2

Simplify

2 a \frac{1}{2} 2 : 2 a

2 a \frac{1}{2} 2

Convert

2

to fraction

: \frac{2}{1}

2

Convert element to fraction:

2 = \frac{2}{1}

= \frac{2}{1}

= 2 a \frac{1}{2} \cdot \frac{2}{1}

Cross-cancel common factor:

2

= 2 a \cdot 1

Apply rule:

a \cdot 1 = a

= 2 a

Simplify

(- 1) 2 : - 2

(- 1) 2

Apply rule:

(- a) = - a

(- 1) = - 1

= - 1 \cdot 2

Apply rule:

1 \cdot a = a

= - 2

2 a = - 2

2 a = - 2

2 a = - 2

Divide both sides by

2

2 a = - 2

Divide both sides by

2

\frac{2 a}{2} = \frac{- 2}{2}

Simplify

\frac{2 a}{2} = \frac{- 2}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Cancel the common factor:

2

= a

Simplify

\frac{- 2}{2} : - \frac{1}{2}

\frac{- 2}{2}

Apply radical rule:

a = a a

2 = 22

= \frac{- 2}{2 2}

Cancel the common factor:

2

= \frac{- 1}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

a = - \frac{1}{2}

For

2 ab = - 1

, subsitute

b

with

- \frac{1}{2} : a = \frac{1}{2}

For

2 ab = - 1

, subsitute

b

with

- \frac{1}{2}

2 a (- \frac{1}{2}) = - 1

Solve

2 a (- \frac{1}{2}) = - 1 : a = \frac{1}{2}

2 a (- \frac{1}{2}) = - 1

Divide both sides by

2 (- \frac{1}{2})

2 a (- \frac{1}{2}) = - 1

Divide both sides by

2 (- \frac{1}{2})

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- 1}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- 1}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : a

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

Simplify

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )}

Apply rule:

a (- b) = - ab

2 a (- \frac{1}{2}) = - 2 a \frac{1}{2}

= \frac{- 2 a \frac{1}{2}}{2 ( - \frac{1}{2} )}

Apply rule:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

= \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

= \frac{- 2 a \frac{1}{2}}{- 2 \cdot \frac{1}{2}}

Cancel the common factor:

- 2

= \frac{a \frac{1}{2}}{\frac{1}{2}}

Cancel the common factor:

\frac{1}{2}

= a

Simplify

\frac{- 1}{2 ( - \frac{1}{2} )} : \frac{1}{2}

\frac{- 1}{2 ( - \frac{1}{2} )}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2 ( - \frac{1}{2} )}

Apply rule:

a (- b) = - ab

2 (- \frac{1}{2}) = - 2 \cdot \frac{1}{2}

= - \frac{1}{- 2 \cdot \frac{1}{2}}

- 2 \cdot \frac{1}{2} = - 2

- 2 \cdot \frac{1}{2}

Convert

2

to fraction

: \frac{2}{1}

2

Convert element to fraction:

2 = \frac{2}{1}

= \frac{2}{1}

= - \frac{2}{1} \cdot \frac{1}{2}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

\frac{2}{1} \cdot \frac{1}{2} = \frac{2 \cdot 1}{1 \cdot 2}

= - \frac{2 \cdot 1}{1 \cdot 2}

\frac{2 \cdot 1}{1 \cdot 2} = 2

\frac{2 \cdot 1}{1 \cdot 2}

\frac{2 \cdot 1}{1 \cdot 2} = \frac{2}{2}

\frac{2 \cdot 1}{1 \cdot 2}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{1 \cdot 2}

Apply rule:

1 \cdot a = a

1 \cdot 2 = 2

= \frac{2}{2}

= \frac{2}{2}

Apply radical rule:

a = a a

2 = 22

= \frac{2 2}{2}

Cancel the common factor:

2

= 2

= - 2

= - \frac{1}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

\frac{1}{- 2} = - \frac{1}{2}

= - (- \frac{1}{2})

Apply rule:

- (- a) = a

- (- \frac{1}{2}) = \frac{1}{2}

= \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

a = \frac{1}{2}

Verify solutions by plugging them into the original equations

Check the solutions by plugging them into

a^{2} - b^{2} = 0

Remove the ones that don't agree with the equation.

Check the solution

a = \frac{1}{2}, b = - \frac{1}{2} :

True

a^{2} - b^{2} = 0

Plug in

a = \frac{1}{2}, b = - \frac{1}{2}

(\frac{1}{2})^{2} - (- \frac{1}{2})^{2} = 0

Refine

0 = 0

True

Check the solution

a = - \frac{1}{2}, b = \frac{1}{2} :

True

a^{2} - b^{2} = 0

Plug in

a = - \frac{1}{2}, b = \frac{1}{2}

(- \frac{1}{2})^{2} - (\frac{1}{2})^{2} = 0

Refine

0 = 0

True

Check the solutions by plugging them into

2 ab = - 1

Remove the ones that don't agree with the equation.

Check the solution

a = \frac{1}{2}, b = - \frac{1}{2} :

True

2 ab = - 1

Plug in

a = \frac{1}{2}, b = - \frac{1}{2}

2 \cdot \frac{1}{2} (- \frac{1}{2}) = - 1

Refine

- 1 = - 1

True

Check the solution

a = - \frac{1}{2}, b = \frac{1}{2} :

True

2 ab = - 1

Plug in

a = - \frac{1}{2}, b = \frac{1}{2}

2 (- \frac{1}{2}) \frac{1}{2} = - 1

Refine

- 1 = - 1

True

Therefore, the final solutions for

a^{2} - b^{2} = 0, 2 ab = - 1

are

(a = - \frac{1}{2}, a = \frac{1}{2}, b = \frac{1}{2} b = - \frac{1}{2})

Substitute back

u = a + bi

u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

The solutions are

u = 0, u = \frac{1}{2} + \frac{1}{2} i, u = - \frac{1}{2} - \frac{1}{2} i, u = - \frac{1}{2} + \frac{1}{2} i, u = \frac{1}{2} - \frac{1}{2} i

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = \frac{1}{2} + \frac{1}{2} i, cos (x) = - \frac{1}{2} - \frac{1}{2} i, cos (x) = - \frac{1}{2} + \frac{1}{2} i, cos (x) = \frac{1}{2} - \frac{1}{2} i

cos (x) = 0, cos (x) = \frac{1}{2} + \frac{1}{2} i, cos (x) = - \frac{1}{2} - \frac{1}{2} i, cos (x) = - \frac{1}{2} + \frac{1}{2} i, cos (x) = \frac{1}{2} - \frac{1}{2} i

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = \frac{1}{2} + \frac{1}{2} i :

No Solution

cos (x) = \frac{1}{2} + \frac{1}{2} i

Simplify

\frac{1}{2} + \frac{1}{2} i : \frac{2}{2} + i \frac{2}{2}

\frac{1}{2} + \frac{1}{2} i

Multiply

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + i}{2}

Rationalize

\frac{1 + i}{2} : \frac{2 ( 1 + i )}{2}

\frac{1 + i}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{( 1 + i ) 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2 ( 1 + i )}{2}

= \frac{2 ( 1 + i )}{2}

Rewrite

\frac{2 ( 1 + i )}{2}

in standard complex form:

\frac{2}{2} + \frac{2}{2} i

\frac{2 ( 1 + i )}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( 1 + i )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

= \frac{1 + i}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{1 + i}{2 ^{\frac{1}{2}}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= \frac{1 + i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 + i}{2} = \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

\frac{1}{2} = \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{1}{2} + \frac{2}{2} i

\frac{1}{2} = \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2} + \frac{2}{2} i

= \frac{2}{2} + \frac{2}{2} i

No Solution

cos (x) = - \frac{1}{2} - \frac{1}{2} i :

No Solution

cos (x) = - \frac{1}{2} - \frac{1}{2} i

Simplify

- \frac{1}{2} - \frac{1}{2} i : - \frac{2}{2} - i \frac{2}{2}

- \frac{1}{2} - \frac{1}{2} i

Multiply

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= - \frac{1}{2} - \frac{i}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 - i}{2}

Rationalize

\frac{- 1 - i}{2} : \frac{2 ( - 1 - i )}{2}

\frac{- 1 - i}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{( - 1 - i ) 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2 ( - 1 - i )}{2}

= \frac{2 ( - 1 - i )}{2}

Rewrite

\frac{2 ( - 1 - i )}{2}

in standard complex form:

- \frac{2}{2} - \frac{2}{2} i

\frac{2 ( - 1 - i )}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( - 1 - i )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

= \frac{- 1 - i}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 - i}{2 ^{\frac{1}{2}}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= \frac{- 1 - i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - i}{2} = - \frac{1}{2} - \frac{i}{2}

= - \frac{1}{2} - \frac{i}{2}

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{1}{2} - \frac{2}{2} i

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2} - \frac{2}{2} i

= - \frac{2}{2} - \frac{2}{2} i

No Solution

cos (x) = - \frac{1}{2} + \frac{1}{2} i :

No Solution

cos (x) = - \frac{1}{2} + \frac{1}{2} i

Simplify

- \frac{1}{2} + \frac{1}{2} i : - \frac{2}{2} + i \frac{2}{2}

- \frac{1}{2} + \frac{1}{2} i

Multiply

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= - \frac{1}{2} + \frac{i}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 + i}{2}

Rationalize

\frac{- 1 + i}{2} : \frac{2 ( - 1 + i )}{2}

\frac{- 1 + i}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{( - 1 + i ) 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2 ( - 1 + i )}{2}

= \frac{2 ( - 1 + i )}{2}

Rewrite

\frac{2 ( - 1 + i )}{2}

in standard complex form:

- \frac{2}{2} + \frac{2}{2} i

\frac{2 ( - 1 + i )}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( - 1 + i )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

= \frac{- 1 + i}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{- 1 + i}{2 ^{\frac{1}{2}}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= \frac{- 1 + i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + i}{2} = - \frac{1}{2} + \frac{i}{2}

= - \frac{1}{2} + \frac{i}{2}

\frac{1}{2} = \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= - \frac{1}{2} + \frac{2}{2} i

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2} + \frac{2}{2} i

= - \frac{2}{2} + \frac{2}{2} i

No Solution

cos (x) = \frac{1}{2} - \frac{1}{2} i :

No Solution

cos (x) = \frac{1}{2} - \frac{1}{2} i

Simplify

\frac{1}{2} - \frac{1}{2} i : \frac{2}{2} - i \frac{2}{2}

\frac{1}{2} - \frac{1}{2} i

Multiply

\frac{1}{2} i : \frac{i}{2}

\frac{1}{2} i

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= \frac{1}{2} - \frac{i}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - i}{2}

Rationalize

\frac{1 - i}{2} : \frac{2 ( 1 - i )}{2}

\frac{1 - i}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{( 1 - i ) 2}{2 2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2 ( 1 - i )}{2}

= \frac{2 ( 1 - i )}{2}

Rewrite

\frac{2 ( 1 - i )}{2}

in standard complex form:

\frac{2}{2} - \frac{2}{2} i

\frac{2 ( 1 - i )}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( 1 - i )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

= \frac{1 - i}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{1 - i}{2 ^{\frac{1}{2}}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= \frac{1 - i}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 - i}{2} = \frac{1}{2} - \frac{i}{2}

= \frac{1}{2} - \frac{i}{2}

- \frac{1}{2} = - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= \frac{1}{2} - \frac{2}{2} i

\frac{1}{2} = \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2} - \frac{2}{2} i

= \frac{2}{2} - \frac{2}{2} i

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos^6(x)=-cos^2(x) ?
The general solution for cos^6(x)=-cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin