Solve for θ cos(2theta)=-( square root of 3)/2

$\cos (2 θ) = - \frac{\sqrt{3}}{2}$

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

$2 θ = \arccos (- \frac{\sqrt{3}}{2})$

Simplify the right side.

The exact value of $\arccos (- \frac{\sqrt{3}}{2})$ is $\frac{5 π}{6}$ .

$2 θ = \frac{5 π}{6}$

Divide each term in $2 θ = \frac{5 π}{6}$ by $2$ and simplify.

Divide each term in $2 θ = \frac{5 π}{6}$ by $2$ .

$\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{5 π}{6}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{5 π}{6} \cdot \frac{1}{2}$

Multiply $\frac{5 π}{6} \cdot \frac{1}{2}$ .

Multiply $\frac{5 π}{6}$ by $\frac{1}{2}$ .

$θ = \frac{5 π}{6 \cdot 2}$

Multiply $6$ by $2$ .

$θ = \frac{5 π}{12}$

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.

$2 θ = 2 π - \frac{5 π}{6}$

Solve for $θ$ .

Simplify.

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 θ = 2 π \cdot \frac{6}{6} - \frac{5 π}{6}$

Combine $2 π$ and $\frac{6}{6}$ .

$2 θ = \frac{2 π \cdot 6}{6} - \frac{5 π}{6}$

Combine the numerators over the common denominator.

$2 θ = \frac{2 π \cdot 6 - 5 π}{6}$

Multiply $6$ by $2$ .

$2 θ = \frac{12 π - 5 π}{6}$

Subtract $5 π$ from $12 π$ .

$2 θ = \frac{7 π}{6}$

Divide each term in $2 θ = \frac{7 π}{6}$ by $2$ and simplify.

Divide each term in $2 θ = \frac{7 π}{6}$ by $2$ .

$\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{7 π}{6}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{7 π}{6} \cdot \frac{1}{2}$

Multiply $\frac{7 π}{6} \cdot \frac{1}{2}$ .

Multiply $\frac{7 π}{6}$ by $\frac{1}{2}$ .

$θ = \frac{7 π}{6 \cdot 2}$

Multiply $6$ by $2$ .

$θ = \frac{7 π}{12}$

Find the period of $\cos (2 θ)$ .

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

$\frac{2 π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

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

$θ = \frac{5 π}{12} + π n, \frac{7 π}{12} + π n$ , for any integer $n$

Solve for θ cos(2theta)=-( square root of 3)/2

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