Find all angles satisfying sin 2θ = | cos θ|

Answer 1

If cosθ ≥ 0 then sin2θ = cosθ; cosθ(2sinθ - 1) = 0, cosθ = 0, θ = π/2 + πn, n = 0, ±1, ±2,...

2sinθ - 1 = 0, sinθ = 1/2. We have to take only solution in 1st quadrant, so θ = π/6 + 2πn, .n = 0, ±1, ±2,...

Now if cosθ < 0, then sin2θ = - cosθ; cosθ(2sinθ + 1) = 0, so 2sinθ + 1 = 0, sinθ = - 1/2,

Now we have to take solution of this equation in quadrant II, where cosθ < 0, so x = - π/6 + 2πn, n = 0, ±1, ±2,...

Answer: π/2 +πn, x = ±π/6 + 2πn, n = 0, ±1, ±2, ±3, ...