Use identity cos 2x = 1 − 2 sin²x. Substitute: 1 − 2 sin²x + sin x = 0 => 2 sin²x − sin x − 1 = 0. Let u = sin x: 2u² − u − 1 = 0 => (2u+1)(u−1)=0 => u = −1/2 or u = 1. So sin x = −1/2 gives x = 7π/6, 11π/6. sin x = 1 gives x = π/2. Solutions: π/2, 7π/6, 11π/6.

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