Solve for θ sin(2theta-pi/2)=-1

$\sin (2 θ - \frac{π}{2}) = - 1$

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

$2 θ - \frac{π}{2} = \arcsin (- 1)$

Simplify the right side.

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

$2 θ - \frac{π}{2} = - \frac{π}{2}$

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

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

$2 θ = - \frac{π}{2} + \frac{π}{2}$

Combine the numerators over the common denominator.

$2 θ = \frac{- π + π}{2}$

Add $- π$ and $π$ .

$2 θ = \frac{0}{2}$

Divide $0$ by $2$ .

$2 θ = 0$

Divide each term in $2 θ = 0$ by $2$ and simplify.

Divide each term in $2 θ = 0$ by $2$ .

$\frac{2 θ}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{0}{2}$

Divide $θ$ by $1$ .

$θ = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$θ = 0$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

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

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

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

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$2 θ - \frac{π}{2} = \frac{3 π}{2}$

Solve for $θ$ .

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

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

$2 θ = \frac{3 π}{2} + \frac{π}{2}$

Combine the numerators over the common denominator.

$2 θ = \frac{3 π + π}{2}$

Add $3 π$ and $π$ .

$2 θ = \frac{4 π}{2}$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4 π$ .

$2 θ = \frac{2 (2 π)}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$2 θ = \frac{2 (2 π)}{2 (1)}$

Cancel the common factor.

$2 θ = \frac{2 (2 π)}{2 \cdot 1}$

Rewrite the expression.

$2 θ = \frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 θ = 2 π$

Divide each term in $2 θ = 2 π$ by $2$ and simplify.

Divide each term in $2 θ = 2 π$ by $2$ .

$\frac{2 θ}{2} = \frac{2 π}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{2 π}{2}$

Divide $θ$ by $1$ .

$θ = \frac{2 π}{2}$

Simplify the right side.

Cancel the common factor of $2$ .

Cancel the common factor.

$θ = \frac{2 π}{2}$

Divide $π$ by $1$ .

$θ = π$

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

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

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

Consolidate the answers.

$θ = π n$ , for any integer $n$

Solve for θ sin(2theta-pi/2)=-1

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