What is the general solution for cos(x)-sqrt(1-3cos^2(x))=0 ?

The general solution for cos(x)-sqrt(1-3cos^2(x))=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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Popular Trigonometry >

cos(x)-sqrt(1-3cos^2(x))=0

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

u - 1 - 3 u^{2} = 0

u - 1 - 3 u^{2} = 0 : u = \frac{1}{2}

u - 1 - 3 u^{2} = 0

Remove square roots

u - 1 - 3 u^{2} = 0

Subtract

u

from both sides

u - 1 - 3 u^{2} - u = 0 - u

Simplify

- 1 - 3 u^{2} = - u

Square both sides:

1 - 3 u^{2} = u^{2}

u - 1 - 3 u^{2} = 0

(- 1 - 3 u^{2})^{2} = (- u)^{2}

Expand

(- 1 - 3 u^{2})^{2} : 1 - 3 u^{2}

(- 1 - 3 u^{2})^{2}

Apply exponent rule:

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

n

is even

(- 1 - 3 u^{2})^{2} = (1 - 3 u^{2})^{2}

= (1 - 3 u^{2})^{2}

Apply radical rule:

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

= ((1 - 3 u^{2})^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (1 - 3 u^{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 - 3 u^{2}

Expand

(- u)^{2} : u^{2}

(- u)^{2}

Apply exponent rule:

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

n

is even

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

= u^{2}

1 - 3 u^{2} = u^{2}

1 - 3 u^{2} = u^{2}

1 - 3 u^{2} = u^{2}

Solve

1 - 3 u^{2} = u^{2} : u = \frac{1}{2}, u = - \frac{1}{2}

1 - 3 u^{2} = u^{2}

Move

1

to the right side

1 - 3 u^{2} = u^{2}

Subtract

1

from both sides

1 - 3 u^{2} - 1 = u^{2} - 1

Simplify

- 3 u^{2} = u^{2} - 1

- 3 u^{2} = u^{2} - 1

Move

u^{2}

to the left side

- 3 u^{2} = u^{2} - 1

Subtract

u^{2}

from both sides

- 3 u^{2} - u^{2} = u^{2} - 1 - u^{2}

Simplify

- 4 u^{2} = - 1

- 4 u^{2} = - 1

Divide both sides by

- 4

- 4 u^{2} = - 1

Divide both sides by

- 4

\frac{- 4 u ^{2}}{- 4} = \frac{- 1}{- 4}

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{4}, u = - \frac{1}{4}

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

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplify

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

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

Verify Solutions:

u = \frac{1}{2}

True

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

False

Check the solutions by plugging them into

u - 1 - 3 u^{2} = 0

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

Plug in

u = \frac{1}{2} :

True

(\frac{1}{2}) - 1 - 3 (\frac{1}{2})^{2} = 0

(\frac{1}{2}) - 1 - 3 (\frac{1}{2})^{2} = 0

(\frac{1}{2}) - 1 - 3 (\frac{1}{2})^{2}

Remove parentheses:

(a) = a

= \frac{1}{2} - 1 - 3 (\frac{1}{2})^{2}

1 - 3 (\frac{1}{2})^{2} = \frac{1}{2}

1 - 3 (\frac{1}{2})^{2}

3 (\frac{1}{2})^{2} = \frac{3}{4}

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= 3 \cdot \frac{1}{2 ^{2}}

Multiply fractions:

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

= \frac{1 \cdot 3}{2 ^{2}}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{2 ^{2}}

2^{2} = 4

= \frac{3}{4}

= 1 - \frac{3}{4}

Join

1 - \frac{3}{4} : \frac{1}{4}

1 - \frac{3}{4}

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 4 - 3}{4}

1 \cdot 4 - 3 = 1

1 \cdot 4 - 3

Multiply the numbers:

1 \cdot 4 = 4

= 4 - 3

Subtract the numbers:

4 - 3 = 1

= 1

= \frac{1}{4}

= \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

Add similar elements:

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

= 0

0 = 0

True

Plug in

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

False

(- \frac{1}{2}) - 1 - 3 (- \frac{1}{2})^{2} = 0

(- \frac{1}{2}) - 1 - 3 (- \frac{1}{2})^{2} = - 1

(- \frac{1}{2}) - 1 - 3 (- \frac{1}{2})^{2}

Remove parentheses:

(- a) = - a

= - \frac{1}{2} - 1 - 3 (- \frac{1}{2})^{2}

1 - 3 (- \frac{1}{2})^{2} = \frac{1}{2}

1 - 3 (- \frac{1}{2})^{2}

3 (- \frac{1}{2})^{2} = \frac{3}{4}

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

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

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

Apply exponent rule:

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

n

is even

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= 3 \cdot \frac{1}{2 ^{2}}

Multiply fractions:

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

= \frac{1 \cdot 3}{2 ^{2}}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{2 ^{2}}

2^{2} = 4

= \frac{3}{4}

= 1 - \frac{3}{4}

Join

1 - \frac{3}{4} : \frac{1}{4}

1 - \frac{3}{4}

Convert element to fraction:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 4 - 3}{4}

1 \cdot 4 - 3 = 1

1 \cdot 4 - 3

Multiply the numbers:

1 \cdot 4 = 4

= 4 - 3

Subtract the numbers:

4 - 3 = 1

= 1

= \frac{1}{4}

= \frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

Apply rule

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

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

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

- 1 = 0

False

The solution is

u = \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)-sqrt(1-3cos^2(x))=0 ?
The general solution for cos(x)-sqrt(1-3cos^2(x))=0 is x= pi/3+2pin,x=(5pi)/3+2pin