Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \) \[ \sin ^{2} x-1=0 \]

Solve the equation \( \sin^{2}(x)-1=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sin^{2}\left(x\right)-1=0\) - step1: Move the constant to the right side: \(\sin^{2}\left(x\right)=0+1\) - step2: Remove 0: \(\sin^{2}\left(x\right)=1\) - step3: Simplify the expression: \(\sin\left(x\right)=\pm \sqrt{1}\) - step4: Simplify: \(\sin\left(x\right)=\pm 1\) - step5: Separate into possible cases: \(\begin{align}&\sin\left(x\right)=1\\&\sin\left(x\right)=-1\end{align}\) - step6: Calculate: \(\begin{align}&x=\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step7: Find the union: \(x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\) The solutions to the trigonometric equation \( \sin^{2}x - 1 = 0 \) for all values \( 0 \leq x < 2\pi \) are \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer.