2cos x - sinx + 1 = 0

Solve 2 cos x − sin x + 1 = 0. A convenient substitution is t = tan(x/2). then - sin x = 2t/(1+t^2) - cos x = (1 − t^2)/(1+t^2) Substitute: 2(1 − t^2)/(1+t^2) − 2t/(1+t^2) + 1 = 0 Multiply by (1+t^2): 2(1 − t^2) − 2t + (1 + t^2) = 0 3 − t^2 − 2t = 0 t^2 + 2t − 3 = 0 (t + 3)(t − 1) = 0 So t = 1 or t = −3. 1) t = tan(x/2) = 1 - x/2 = π/4 + kπ ⇒ x = π/2 + 2kπ 2) t = tan(x/2) = −3 - x/2 = arctan(−3) + kπ ⇒ x = 2 arctan(−3) + 2kπ - since arctan(−3) = −arctan(3), this is x = −2 arctan(3) + 2kπ - numerically, x ≈ −2.49809 + 2kπ, i.e. in [0, 2π): x ≈ 3.78509 rad (216.87°) General solutions: - x = π/2 + 2kπ - x = −2 arctan(3) + 2kπ for any integer k.