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 ± \(\frac{\pi}{6}\)
