The solution to the trigonometric equation \( \frac{1}{2} \csc ^{2} x-1=0 \) is \( x=\longrightarrow x= \) \( x=\longrightarrow \) and \( x=\longrightarrow \) Show your work.

To solve the trigonometric equation: \[ \frac{1}{2} \csc^{2} x - 1 = 0 \] Let's follow the steps to find all solutions for \( x \). ### Step 1: Rewrite the Equation in Terms of Sine Recall that the cosecant function is the reciprocal of the sine function: \[ \csc x = \frac{1}{\sin x} \quad \text{and} \quad \csc^{2} x = \frac{1}{\sin^{2} x} \] Substitute \(\csc^{2} x\) into the equation: \[ \frac{1}{2} \cdot \frac{1}{\sin^{2} x} - 1 = 0 \] ### Step 2: Solve for \(\sin^{2} x\) Multiply both sides of the equation by \(2\) to eliminate the fraction: \[ \frac{1}{\sin^{2} x} - 2 = 0 \] Add \(2\) to both sides: \[ \frac{1}{\sin^{2} x} = 2 \] Take the reciprocal of both sides: \[ \sin^{2} x = \frac{1}{2} \] ### Step 3: Find \(\sin x\) Take the square root of both sides, remembering to consider both positive and negative roots: \[ \sin x = \pm \frac{\sqrt{2}}{2} \] ### Step 4: Determine the General Solutions for \( x \) The equation \(\sin x = \pm \frac{\sqrt{2}}{2}\) has solutions at specific angles within the unit circle. Specifically: 1. **For \(\sin x = \frac{\sqrt{2}}{2}\):** - \( x = \frac{\pi}{4} + 2k\pi \) - \( x = \frac{3\pi}{4} + 2k\pi \) 2. **For \(\sin x = -\frac{\sqrt{2}}{2}\):** - \( x = \frac{5\pi}{4} + 2k\pi \) - \( x = \frac{7\pi}{4} + 2k\pi \) where \( k \) is any integer. ### Step 5: Express the Solutions Concisely Notice that all solutions can be expressed in a unified form by recognizing the periodicity of the sine function (which has a period of \( 2\pi \)) and the symmetry in the unit circle: \[ x = \frac{\pi}{4} + \frac{\pi}{2}k \quad \text{for any integer } k \] This captures all four cases by varying \( k \). However, to present the primary solutions within one period (\( 0 \) to \( 2\pi \)), we can write: \[ \boxed{ \begin{aligned} x &= \frac{\pi}{4} + 2k\pi \\ \text{and} \quad x &= \frac{3\pi}{4} + 2k\pi \\ \end{aligned} \quad \text{for any integer } k } \] ### Final Answer The solutions to the equation \( \frac{1}{2} \csc^{2} x - 1 = 0 \) are: \[ \boxed{ \begin{aligned} x &= \frac{\pi}{4} + 2k\pi \\ \text{and} \quad x &= \frac{3\pi}{4} + 2k\pi \\ \end{aligned} \quad \text{for any integer } k } \]