Solve for θ cos(2theta-pi/2)=-1

$\cos (2 θ - \frac{π}{2}) = - 1$

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

$2 θ - \frac{π}{2} = \arccos (- 1)$

Simplify the right side.

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

$2 θ - \frac{π}{2} = π$

Move all terms not containing $θ$ to the right side of the equation.

Add $\frac{π}{2}$ to both sides of the equation.

$2 θ = π + \frac{π}{2}$

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

$2 θ = π \cdot \frac{2}{2} + \frac{π}{2}$

Combine $π$ and $\frac{2}{2}$ .

$2 θ = \frac{π \cdot 2}{2} + \frac{π}{2}$

Combine the numerators over the common denominator.

$2 θ = \frac{π \cdot 2 + π}{2}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$2 θ = \frac{2 \cdot π + π}{2}$

Add $2 π$ and $π$ .

$2 θ = \frac{3 π}{2}$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $θ$ by $1$ .

$θ = \frac{\frac{3 π}{2}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{3 π}{2} \cdot \frac{1}{2}$

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

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

$θ = \frac{3 π}{2 \cdot 2}$

Multiply $2$ by $2$ .

$θ = \frac{3 π}{4}$

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.

$2 θ - \frac{π}{2} = 2 π - π$

Solve for $θ$ .

Subtract $π$ from $2 π$ .

$2 θ - \frac{π}{2} = π$

Move all terms not containing $θ$ to the right side of the equation.

Add $\frac{π}{2}$ to both sides of the equation.

$2 θ = π + \frac{π}{2}$

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

$2 θ = π \cdot \frac{2}{2} + \frac{π}{2}$

Combine $π$ and $\frac{2}{2}$ .

$2 θ = \frac{π \cdot 2}{2} + \frac{π}{2}$

Combine the numerators over the common denominator.

$2 θ = \frac{π \cdot 2 + π}{2}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$2 θ = \frac{2 \cdot π + π}{2}$

Add $2 π$ and $π$ .

$2 θ = \frac{3 π}{2}$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $θ$ by $1$ .

$θ = \frac{\frac{3 π}{2}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{3 π}{2} \cdot \frac{1}{2}$

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

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

$θ = \frac{3 π}{2 \cdot 2}$

Multiply $2$ by $2$ .

$θ = \frac{3 π}{4}$

Find the period of $\cos (2 θ - \frac{π}{2})$ .

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 $\cos (2 θ - \frac{π}{2})$ function is $π$ so values will repeat every $π$ radians in both directions.

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

Solve for θ cos(2theta-pi/2)=-1

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