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Solve: 2 cos2x + 3 sinx = 0 |
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Answer» We have: 2 cos2x + 3 sinx = 0 2 cos2x= – 3 sinx Squaring both sides, we get 4 cos4x = 9 sin2x 4 cos4x – 9 (1 – cos2x) = 0 Let cos2x = y 4y2 + 9y – 9 = 0 4y2 + 12y – 3y – 9 = 0 4y (y + 3) – 3 (y + 3) = 0 (y + 3) (4y – 3) = 0 y = – 3 or, y = \(\frac{3}{4}\) cos2x = – 3 or cos2x = \(\frac{3}{4}\) ∴ cos2x = \(\frac{3}{4}\) = cos2\((\frac{\pi}{6})\) x = nπ ± \(\frac{\pi}{6}\) |
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