Find all solutions in the interval \( [0,2 \pi) \). \( \sin 2 x \sin x-\cos x=0 \) \( x=\square \) (Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed. Use integers or fractions for any numbers in the expression.) 0 of 1

Solve the equation \( \sin(2x)\sin(x)-\cos(x)=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sin\left(2x\right)\sin\left(x\right)-\cos\left(x\right)=0\) - step1: Rewrite the expression: \(2\sin\left(x\right)\cos\left(x\right)\sin\left(x\right)-\cos\left(x\right)=0\) - step2: Expand the expression: \(2\sin^{2}\left(x\right)\cos\left(x\right)-\cos\left(x\right)=0\) - step3: Factor the expression: \(\cos\left(x\right)\left(2\sin^{2}\left(x\right)-1\right)=0\) - step4: Separate into possible cases: \(\begin{align}&\cos\left(x\right)=0\\&2\sin^{2}\left(x\right)-1=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\\&x=\frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\end{align}\) - step6: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{4}+\frac{k\pi }{2}\\\frac{\pi }{2}+k\pi \end{array}\right.,k \in \mathbb{Z}\) The solutions to the equation \( \sin 2x \sin x - \cos x = 0 \) in the interval \( [0, 2\pi) \) are \( x = \frac{\pi}{4} + \frac{k\pi}{2} \) and \( x = \frac{\pi}{2} + k\pi \) for any integer \( k \).