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To solve the equation \( \sin 3x = \cos 2x \), we'll use trigonometric identities. We know that: \[ \sin 3x = \sin(2x + x) = \sin 2x \cos x + \cos 2x \sin x \] \[ \cos 2x = \cos^2 x - \sin^2 x \] Thus, the equation becomes: \[ \sin 2x \cos x + \cos 2x \sin x = \cos 2x \] Substituting for \(\sin 2x\) and \(\cos 2x\): \[ 2\sin x \cos x \cos x + (\cos^2 x - \sin^2 x) \sin x = \cos^2 x - \sin^2 x \] To solve: 1. Simplify and factor: \[ 2\sin x \cos^2 x + \sin x \cos^2 x - \sin^3 x = \cos^2 x - \sin^2 x \] 2. After simplifying and rearranging terms, use identities and algebra to solve for \(x\). Alternatively, solve \(\tan 3x = 2\) directly or evaluate the initial equation for specific angle multiples since \(\tan(x) = \tan(\pi + x)\). Final solution requires specific calculation or graphing to find \(x\) such that \(3x\) and \(2x\) satisfy the angle condition for tangent equivalence.