Statement Solve the trigonometric equation: 2sin2x+sinx−1=02\sin^2 x + \sin x - 1 = 02sin2x+sinx−1=0 Solution Substitution: t=sinxt = \sin xt=sinx, with −1≤t≤1-1\le t\le 1−1≤t≤1. The equation becomes 2t2+t−1=0 ⟺ (2t−1)(t+1)=0 ⟺ t=12 ∨ t=−1.2t^2 + t - 1 = 0 \iff (2t-1)(t+1) = 0 \iff t = \tfrac{1}{2} \ \vee\ t = -1.2t2+t−1=0⟺(2t−1)(t+1)=0⟺t=21 ∨ t=−1. Both solutions are acceptable. Back to xxx: sinx=12\sin x = \tfrac{1}{2}sinx=21: x=π6+2kπ\ x = \dfrac{\pi}{6} + 2k\pi x=6π+2kπ or x=5π6+2kπx = \dfrac{5\pi}{6} + 2k\pix=65π+2kπ; sinx=−1\sin x = -1sinx=−1: x=−π2+2kπ\ x = -\dfrac{\pi}{2} + 2k\pi x=−2π+2kπ. x=π6+2kπ ∨ x=5π6+2kπ ∨ x=−π2+2kπ,k∈Z.\boxed{\,x = \frac{\pi}{6} + 2k\pi \ \vee\ x = \frac{5\pi}{6} + 2k\pi \ \vee\ x = -\frac{\pi}{2} + 2k\pi,\quad k\in\mathbb{Z}.}x=6π+2kπ ∨ x=65π+2kπ ∨ x=−2π+2kπ,k∈Z. Links Topics: Trigonometric equations Concepts: Elementary equation · Substitution Functions: Sine Skills: Solving equations · Using formulae Exercise type: Solving an equation