What is the general solution for cos^2(x)+cos^2(3x)=1 ?

The general solution for cos^2(x)+cos^2(3x)=1 is

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Popular Trigonometry >

cos^2(x)+cos^2(3x)=1

Solution

cos^{2} (x) + cos^{2} (3 x) = 1

Solution

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn, x = 0.39269 \dots + 2 πn, x = 2 π - 0.39269 \dots + 2 πn, x = 2.74889 \dots + 2 πn, x = - 2.74889 \dots + 2 πn, x = 1.17809 \dots + 2 πn, x = 2 π - 1.17809 \dots + 2 πn, x = 1.96349 \dots + 2 πn, x = - 1.96349 \dots + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n, x = 22. 5^{\circ} + 36 0^{\circ} n, x = 337. 5^{\circ} + 36 0^{\circ} n, x = 157. 5^{\circ} + 36 0^{\circ} n, x = - 157. 5^{\circ} + 36 0^{\circ} n, x = 67. 5^{\circ} + 36 0^{\circ} n, x = 292. 5^{\circ} + 36 0^{\circ} n, x = 112. 5^{\circ} + 36 0^{\circ} n, x = - 112. 5^{\circ} + 36 0^{\circ} n

Solution steps

cos^{2} (x) + cos^{2} (3 x) = 1

Subtract

1

from both sides

cos^{2} (x) + cos^{2} (3 x) - 1 = 0

Rewrite using trig identities

- 1 + cos^{2} (3 x) + cos^{2} (x)

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

Rewrite using trig identities

cos (3 x)

Rewrite as

= cos (2 x + x)

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 x) cos (x) - sin (2 x) sin (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

Simplify

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x) : cos (x) cos (2 x) - 2 sin^{2} (x) cos (x)

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

2 sin (x) cos (x) sin (x) = 2 sin^{2} (x) cos (x)

2 sin (x) cos (x) sin (x)

Apply exponent rule:

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

sin (x) sin (x) = sin^{1 + 1} (x)

= 2 cos (x) sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 2 cos (x) sin^{2} (x)

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

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

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

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= (2 cos^{2} (x) - 1) cos (x) - 2 sin^{2} (x) cos (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= (2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x)

Expand

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x)

= cos (x) (2 cos^{2} (x) - 1) - 2 cos (x) (1 - cos^{2} (x))

Expand

cos (x) (2 cos^{2} (x) - 1) : 2 cos^{3} (x) - cos (x)

cos (x) (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = cos (x), b = 2 cos^{2} (x), c = 1

= cos (x) 2 cos^{2} (x) - cos (x) 1

= 2 cos^{2} (x) cos (x) - 1 cos (x)

Simplify

2 cos^{2} (x) cos (x) - 1 \cdot cos (x) : 2 cos^{3} (x) - cos (x)

2 cos^{2} (x) cos (x) - 1 cos (x)

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

2 cos^{2} (x) cos (x)

Apply exponent rule:

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

cos^{2} (x) cos (x) = cos^{2 + 1} (x)

= 2 cos^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

= 2 cos^{3} (x) - cos (x)

= 2 cos^{3} (x) - cos (x)

= 2 cos^{3} (x) - cos (x) - 2 (1 - cos^{2} (x)) cos (x)

Expand

- 2 cos (x) (1 - cos^{2} (x)) : - 2 cos (x) + 2 cos^{3} (x)

- 2 cos (x) (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2 cos (x), b = 1, c = cos^{2} (x)

= - 2 cos (x) 1 - (- 2 cos (x)) cos^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 cos (x) + 2 cos^{2} (x) cos (x)

Simplify

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x) : - 2 cos (x) + 2 cos^{3} (x)

- 2 \cdot 1 cos (x) + 2 cos^{2} (x) cos (x)

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (x)

2 cos^{2} (x) cos (x) = 2 cos^{3} (x)

2 cos^{2} (x) cos (x)

Apply exponent rule:

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

cos^{2} (x) cos (x) = cos^{2 + 1} (x)

= 2 cos^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= 2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Simplify

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x) : 4 cos^{3} (x) - 3 cos (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Group like terms

= 2 cos^{3} (x) + 2 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

- cos (x) - 2 cos (x) = - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= - 1 + (4 cos^{3} (x) - 3 cos (x))^{2} + cos^{2} (x)

Simplify

- 1 + (4 cos^{3} (x) - 3 cos (x))^{2} + cos^{2} (x) : - 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 10 cos^{2} (x)

- 1 + (4 cos^{3} (x) - 3 cos (x))^{2} + cos^{2} (x)

(4 cos^{3} (x) - 3 cos (x))^{2} : 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

Apply Perfect Square Formula:

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

a = 4 cos^{3} (x), b = 3 cos (x)

= (4 cos^{3} (x))^{2} - 2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) + (3 cos (x))^{2}

Simplify

(4 cos^{3} (x))^{2} - 2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) + (3 cos (x))^{2} : 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

(4 cos^{3} (x))^{2} - 2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) + (3 cos (x))^{2}

(4 cos^{3} (x))^{2} = 16 cos^{6} (x)

(4 cos^{3} (x))^{2}

Apply exponent rule:

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

= 4^{2} (cos^{3} (x))^{2}

(cos^{3} (x))^{2} : cos^{6} (x)

Apply exponent rule:

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

= cos^{3 \cdot 2} (x)

Multiply the numbers:

3 \cdot 2 = 6

= cos^{6} (x)

= 4^{2} cos^{6} (x)

4^{2} = 16

= 16 cos^{6} (x)

2 \cdot 4 cos^{3} (x) \cdot 3 cos (x) = 24 cos^{4} (x)

2 \cdot 4 cos^{3} (x) \cdot 3 cos (x)

Multiply the numbers:

2 \cdot 4 \cdot 3 = 24

= 24 cos^{3} (x) cos (x)

Apply exponent rule:

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

cos^{3} (x) cos (x) = cos^{3 + 1} (x)

= 24 cos^{3 + 1} (x)

Add the numbers:

3 + 1 = 4

= 24 cos^{4} (x)

(3 cos (x))^{2} = 9 cos^{2} (x)

(3 cos (x))^{2}

Apply exponent rule:

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

= 3^{2} cos^{2} (x)

3^{2} = 9

= 9 cos^{2} (x)

= 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

= 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x)

= - 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 9 cos^{2} (x) + cos^{2} (x)

Add similar elements:

9 cos^{2} (x) + cos^{2} (x) = 10 cos^{2} (x)

= - 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 10 cos^{2} (x)

= - 1 + 16 cos^{6} (x) - 24 cos^{4} (x) + 10 cos^{2} (x)

- 1 + 10 cos^{2} (x) + 16 cos^{6} (x) - 24 cos^{4} (x) = 0

Solve by substitution

- 1 + 10 cos^{2} (x) + 16 cos^{6} (x) - 24 cos^{4} (x) = 0

Let:

cos (x) = u

- 1 + 10 u^{2} + 16 u^{6} - 24 u^{4} = 0

- 1 + 10 u^{2} + 16 u^{6} - 24 u^{4} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}, u = \frac{2 + 2}{2}, u = - \frac{2 + 2}{2}, u = \frac{2 - 2}{2}, u = - \frac{2 - 2}{2}

- 1 + 10 u^{2} + 16 u^{6} - 24 u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

16 u^{6} - 24 u^{4} + 10 u^{2} - 1 = 0

Rewrite the equation with

v = u^{2}, v^{2} = u^{4}

and

v^{3} = u^{6}

16 v^{3} - 24 v^{2} + 10 v - 1 = 0

Solve

16 v^{3} - 24 v^{2} + 10 v - 1 = 0 : v = \frac{1}{2}, v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

16 v^{3} - 24 v^{2} + 10 v - 1 = 0

Factor

16 v^{3} - 24 v^{2} + 10 v - 1 : (2 v - 1) (8 v^{2} - 8 v + 1)

16 v^{3} - 24 v^{2} + 10 v - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 16

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4, 8, 16

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4 , 8 , 16}

\frac{1}{2}

is a root of the expression, so factor out

2 v - 1

= (2 v - 1) \frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1}

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1} = 8 v^{2} - 8 v + 1

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1}

Divide

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1} : \frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1} = 8 v^{2} + \frac{- 16 v ^{2} + 10 v - 1}{2 v - 1}

Divide the leading coefficients of the numerator

16 v^{3} - 24 v^{2} + 10 v - 1

and the divisor

2 v - 1 : \frac{16 v ^{3}}{2 v} = 8 v^{2}

Quotient = 8 v^{2}

Multiply

2 v - 1

8 v^{2} : 16 v^{3} - 8 v^{2}

Subtract

16 v^{3} - 8 v^{2}

from

16 v^{3} - 24 v^{2} + 10 v - 1

to get new remainder

Remainder = - 16 v^{2} + 10 v - 1

Therefore

\frac{16 v ^{3} - 24 v ^{2} + 10 v - 1}{2 v - 1} = 8 v^{2} + \frac{- 16 v ^{2} + 10 v - 1}{2 v - 1}

= 8 v^{2} + \frac{- 16 v ^{2} + 10 v - 1}{2 v - 1}

Divide

\frac{- 16 v ^{2} + 10 v - 1}{2 v - 1} : \frac{- 16 v ^{2} + 10 v - 1}{2 v - 1} = - 8 v + \frac{2 v - 1}{2 v - 1}

Divide the leading coefficients of the numerator

- 16 v^{2} + 10 v - 1

and the divisor

2 v - 1 : \frac{- 16 v ^{2}}{2 v} = - 8 v

Quotient = - 8 v

Multiply

2 v - 1

- 8 v : - 16 v^{2} + 8 v

Subtract

- 16 v^{2} + 8 v

from

- 16 v^{2} + 10 v - 1

to get new remainder

Remainder = 2 v - 1

Therefore

\frac{- 16 v ^{2} + 10 v - 1}{2 v - 1} = - 8 v + \frac{2 v - 1}{2 v - 1}

= 8 v^{2} - 8 v + \frac{2 v - 1}{2 v - 1}

Divide

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

Divide the leading coefficients of the numerator

2 v - 1

and the divisor

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

Quotient = 1

Multiply

2 v - 1

1 : 2 v - 1

Subtract

2 v - 1

from

2 v - 1

to get new remainder

Remainder = 0

Therefore

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

= 8 v^{2} - 8 v + 1

= (2 v - 1) (8 v^{2} - 8 v + 1)

(2 v - 1) (8 v^{2} - 8 v + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 v - 1 = 0 or 8 v^{2} - 8 v + 1 = 0

Solve

2 v - 1 = 0 : v = \frac{1}{2}

2 v - 1 = 0

Move

1

to the right side

2 v - 1 = 0

Add

1

to both sides

2 v - 1 + 1 = 0 + 1

Simplify

2 v = 1

2 v = 1

Divide both sides by

2

2 v = 1

Divide both sides by

2

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

Simplify

v = \frac{1}{2}

v = \frac{1}{2}

Solve

8 v^{2} - 8 v + 1 = 0 : v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

8 v^{2} - 8 v + 1 = 0

Solve with the quadratic formula

8 v^{2} - 8 v + 1 = 0

Quadratic Equation Formula:

For

a = 8, b = - 8, c = 1

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 8 \cdot 1}{2 \cdot 8}

(- 8)^{2} - 4 \cdot 8 \cdot 1 = 42

(- 8)^{2} - 4 \cdot 8 \cdot 1

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 8 \cdot 1

Multiply the numbers:

4 \cdot 8 \cdot 1 = 32

= 8^{2} - 32

8^{2} = 64

= 64 - 32

Subtract the numbers:

64 - 32 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

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

= 2^{4} \cdot 2

Apply radical rule:

= 2 2^{4}

Apply radical rule:

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

= 2^{2} 2

Refine

= 42

v_{1, 2} = \frac{- ( - 8 ) \pm 4 2}{2 \cdot 8}

Separate the solutions

v_{1} = \frac{- ( - 8 ) + 4 2}{2 \cdot 8}, v_{2} = \frac{- ( - 8 ) - 4 2}{2 \cdot 8}

v = \frac{- ( - 8 ) + 4 2}{2 \cdot 8} : \frac{2 + 2}{4}

\frac{- ( - 8 ) + 4 2}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{8 + 4 2}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8 + 4 2}{16}

Factor

8 + 42 : 4 (2 + 2)

8 + 42

Rewrite as

= 4 \cdot 2 + 42

Factor out common term

4

= 4 (2 + 2)

= \frac{4 ( 2 + 2 )}{16}

Cancel the common factor:

4

= \frac{2 + 2}{4}

v = \frac{- ( - 8 ) - 4 2}{2 \cdot 8} : \frac{2 - 2}{4}

\frac{- ( - 8 ) - 4 2}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{8 - 4 2}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8 - 4 2}{16}

Factor

8 - 42 : 4 (2 - 2)

8 - 42

Rewrite as

= 4 \cdot 2 - 42

Factor out common term

4

= 4 (2 - 2)

= \frac{4 ( 2 - 2 )}{16}

Cancel the common factor:

4

= \frac{2 - 2}{4}

The solutions to the quadratic equation are:

v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

The solutions are

v = \frac{1}{2}, v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

v = \frac{1}{2}, v = \frac{2 + 2}{4}, v = \frac{2 - 2}{4}

Substitute back

v = u^{2},

solve for

u

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

Solve

u^{2} = \frac{2 + 2}{4} : u = \frac{2 + 2}{2}, u = - \frac{2 + 2}{2}

u^{2} = \frac{2 + 2}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{2 + 2}{4}, u = - \frac{2 + 2}{4}

\frac{2 + 2}{4} = \frac{2 + 2}{2}

\frac{2 + 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2 + 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2 + 2}{2}

- \frac{2 + 2}{4} = - \frac{2 + 2}{2}

- \frac{2 + 2}{4}

Simplify

\frac{2 + 2}{4} : \frac{2 + 2}{2}

\frac{2 + 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2 + 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2 + 2}{2}

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

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

Solve

u^{2} = \frac{2 - 2}{4} : u = \frac{2 - 2}{2}, u = - \frac{2 - 2}{2}

u^{2} = \frac{2 - 2}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{2 - 2}{4}, u = - \frac{2 - 2}{4}

\frac{2 - 2}{4} = \frac{2 - 2}{2}

\frac{2 - 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2 - 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2 - 2}{2}

- \frac{2 - 2}{4} = - \frac{2 - 2}{2}

- \frac{2 - 2}{4}

Simplify

\frac{2 - 2}{4} : \frac{2 - 2}{2}

\frac{2 - 2}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{2 - 2}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2 - 2}{2}

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

u = \frac{2 - 2}{2}, u = - \frac{2 - 2}{2}

The solutions are

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

cos (x) = \frac{2 + 2}{2} : x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn

x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn

cos (x) = - \frac{2 + 2}{2} : x = arccos (- \frac{2 + 2}{2}) + 2 πn, x = - arccos (- \frac{2 + 2}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{2 + 2}{2}) + 2 πn, x = - arccos (- \frac{2 + 2}{2}) + 2 πn

x = arccos (- \frac{2 + 2}{2}) + 2 πn, x = - arccos (- \frac{2 + 2}{2}) + 2 πn

cos (x) = \frac{2 - 2}{2} : x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn

x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn

cos (x) = - \frac{2 - 2}{2} : x = arccos (- \frac{2 - 2}{2}) + 2 πn, x = - arccos (- \frac{2 - 2}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{2 - 2}{2}) + 2 πn, x = - arccos (- \frac{2 - 2}{2}) + 2 πn

x = arccos (- \frac{2 - 2}{2}) + 2 πn, x = - arccos (- \frac{2 - 2}{2}) + 2 πn

Combine all the solutions

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn, x = arccos (\frac{2 + 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 + 2}{2}) + 2 πn, x = arccos (- \frac{2 + 2}{2}) + 2 πn, x = - arccos (- \frac{2 + 2}{2}) + 2 πn, x = arccos (\frac{2 - 2}{2}) + 2 πn, x = 2 π - arccos (\frac{2 - 2}{2}) + 2 πn, x = arccos (- \frac{2 - 2}{2}) + 2 πn, x = - arccos (- \frac{2 - 2}{2}) + 2 πn

Show solutions in decimal form

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn, x = 0.39269 \dots + 2 πn, x = 2 π - 0.39269 \dots + 2 πn, x = 2.74889 \dots + 2 πn, x = - 2.74889 \dots + 2 πn, x = 1.17809 \dots + 2 πn, x = 2 π - 1.17809 \dots + 2 πn, x = 1.96349 \dots + 2 πn, x = - 1.96349 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)+cos^2(3x)=1 ?
The general solution for cos^2(x)+cos^2(3x)=1 is