solutions for \( \sin 2 x=\sin x, 0 \leq x \leq 2 \pi \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\sin\left(2x\right)-\sin\left(x\right)=0\) - step1: Transform the expression: \(2\cos\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)=0\) - step2: Elimination the left coefficient: \(\cos\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)=0\) - step3: Separate into possible cases: \(\begin{align}&\cos\left(\frac{3x}{2}\right)=0\\&\sin\left(\frac{x}{2}\right)=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=\frac{\pi }{3}+\frac{2k\pi }{3},k \in \mathbb{Z}\\&x=2k\pi ,k \in \mathbb{Z}\end{align}\) - step5: Find the union: \(x=\left\{ \begin{array}{l}2k\pi \\\frac{\pi }{3}+\frac{2k\pi }{3}\end{array}\right.,k \in \mathbb{Z}\) To solve the equation \( \sin(2x) = \sin(x) \) within the interval \( 0 \leq x \leq 2\pi \), we can start by rewriting the equation: \[ \sin(2x) - \sin(x) = 0 \] Using the double angle identity, we know that \( \sin(2x) = 2\sin(x)\cos(x) \). Thus, we can rewrite the equation as: \[ 2\sin(x)\cos(x) - \sin(x) = 0 \] Factoring out \( \sin(x) \): \[ \sin(x)(2\cos(x) - 1) = 0 \] This gives us two cases to solve: 1. \( \sin(x) = 0 \) 2. \( 2\cos(x) - 1 = 0 \) ### Case 1: \( \sin(x) = 0 \) The solutions for \( \sin(x) = 0 \) in the interval \( 0 \leq x \leq 2\pi \) are: \[ x = 0, \pi, 2\pi \] ### Case 2: \( 2\cos(x) - 1 = 0 \) Solving for \( \cos(x) \): \[ \cos(x) = \frac{1}{2} \] The solutions for \( \cos(x) = \frac{1}{2} \) in the interval \( 0 \leq x \leq 2\pi \) are: \[ x = \frac{\pi}{3}, \frac{5\pi}{3} \] ### Final Solutions Combining the solutions from both cases, we have: \[ x = 0, \pi, 2\pi, \frac{\pi}{3}, \frac{5\pi}{3} \] Thus, the complete set of solutions for \( \sin(2x) = \sin(x) \) in the interval \( 0 \leq x \leq 2\pi \) is: \[ x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi \]