2cos²x+5sin-4=0

Solve 2 cos^2 x + 5 sin x - 4 = 0.

• Use cos^2 x = 1 − sin^2 x. Let s = sin x.
• Then 2(1 − s^2) + 5s − 4 = 0 → −2s^2 + 5s − 2 = 0 → 2s^2 − 5s + 2 = 0.
• Solve: s = [5 ± √(25 − 16)]/4 = (5 ± 3)/4 → s = 2 or s = 1/2.
• sin x = 2 is impossible, so sin x = 1/2.

Thus x = π/6 + 2kπ or x = 5π/6 + 2kπ, where k ∈ Z. If restricting to [0, 2π): x = π/6 or x = 5π/6.