Solve for θ cos(2theta)=cos(theta)

$\cos (2 θ) = \cos (θ)$

Subtract $\cos (θ)$ from both sides of the equation.

$\cos (2 θ) - \cos (θ) = 0$

Use the double-angle identity to transform $\cos (2 x)$ to $2 \cos^{2} (x) - 1$ .

$2 \cos^{2} (θ) - 1 - \cos (θ) = 0$

Factor by grouping.

Reorder terms.

$2 \cos^{2} (θ) - \cos (θ) - 1 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- \cos (θ)$ .

$2 \cos^{2} (θ) - \cos (θ) - 1 = 0$

Rewrite $- 1$ as $1$ plus $- 2$

$2 \cos^{2} (θ) + (1 - 2) \cos (θ) - 1 = 0$

Apply the distributive property.

$2 \cos^{2} (θ) + 1 \cos (θ) - 2 \cos (θ) - 1 = 0$

Multiply $\cos (θ)$ by $1$ .

$2 \cos^{2} (θ) + \cos (θ) - 2 \cos (θ) - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 \cos^{2} (θ) + \cos (θ) - 2 \cos (θ) - 1 = 0$

Factor out the greatest common factor (GCF) from each group.

$\cos (θ) (2 \cos (θ) + 1) - (2 \cos (θ) + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 \cos (θ) + 1$ .

$(2 \cos (θ) + 1) (\cos (θ) - 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 \cos (θ) + 1 = 0$

$\cos (θ) - 1 = 0$

Set $2 \cos (θ) + 1$ equal to $0$ and solve for $θ$ .

Set $2 \cos (θ) + 1$ equal to $0$ .

$2 \cos (θ) + 1 = 0$

Solve $2 \cos (θ) + 1 = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$2 \cos (θ) = - 1$

Divide each term in $2 \cos (θ) = - 1$ by $2$ and simplify.

Divide each term in $2 \cos (θ) = - 1$ by $2$ .

$\frac{2 \cos (θ)}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \cos (θ)}{2} = \frac{- 1}{2}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$\cos (θ) = - \frac{1}{2}$

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (- \frac{1}{2})$

Simplify the right side.

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$θ = \frac{2 π}{3}$

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$θ = 2 π - \frac{2 π}{3}$

Simplify $2 π - \frac{2 π}{3}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$θ = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$θ = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$θ = \frac{4 π}{3}$

Find the period of $\cos (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Set $\cos (θ) - 1$ equal to $0$ and solve for $θ$ .

Set $\cos (θ) - 1$ equal to $0$ .

$\cos (θ) - 1 = 0$

Solve $\cos (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$\cos (θ) = 1$

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (1)$

Simplify the right side.

The exact value of $\arccos (1)$ is $0$ .

$θ = 0$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - 0$

Subtract $0$ from $2 π$ .

$θ = 2 π$

Find the period of $\cos (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = 2 π n, 2 π + 2 π n$ , for any integer $n$

The final solution is all the values that make $(2 \cos (θ) + 1) (\cos (θ) - 1) = 0$ true.

$θ = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, 2 π n, 2 π + 2 π n$ , for any integer $n$

Consolidate the answers.

$θ = \frac{2 π n}{3}$ , for any integer $n$

Solve for θ cos(2theta)=cos(theta)

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