Solve for θ 2cos(2theta)-1=0

$2 \cos (2 θ) - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$2 \cos (2 θ) = 1$

Step 2

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $\cos (2 θ)$ by $1$ .

$\cos (2 θ) = \frac{1}{2}$

Step 3

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

$2 θ = \arccos (\frac{1}{2})$

Step 4

Simplify the right side.

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

$2 θ = \frac{π}{3}$

Step 5

Divide each term in $2 θ = \frac{π}{3}$ by $2$ and simplify.

Divide each term in $2 θ = \frac{π}{3}$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $θ$ by $1$ .

$θ = \frac{\frac{π}{3}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{π}{3} \cdot \frac{1}{2}$

Multiply $\frac{π}{3} \cdot \frac{1}{2}$ .

Multiply $\frac{π}{3}$ by $\frac{1}{2}$ .

$θ = \frac{π}{3 \cdot 2}$

Multiply $3$ by $2$ .

$θ = \frac{π}{6}$

Step 6

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.

$2 θ = 2 π - \frac{π}{3}$

Step 7

Solve for $θ$ .

Simplify.

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

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

Combine $2 π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$2 θ = \frac{2 π \cdot 3 - π}{3}$

Multiply $3$ by $2$ .

$2 θ = \frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$2 θ = \frac{5 π}{3}$

Divide each term in $2 θ = \frac{5 π}{3}$ by $2$ and simplify.

Divide each term in $2 θ = \frac{5 π}{3}$ by $2$ .

$\frac{2 θ}{2} = \frac{\frac{5 π}{3}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{5 π}{3}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{5 π}{3}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{5 π}{3} \cdot \frac{1}{2}$

Multiply $\frac{5 π}{3} \cdot \frac{1}{2}$ .

Multiply $\frac{5 π}{3}$ by $\frac{1}{2}$ .

$θ = \frac{5 π}{3 \cdot 2}$

Multiply $3$ by $2$ .

$θ = \frac{5 π}{6}$

Step 8

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 $1$ .

$π$

Step 9

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

$θ = \frac{π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Solve for θ 2cos(2theta)-1=0

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