Solve the equation \(2\sin^2x - 3\sin x - 2 = 0\) in the interval \([0, 2\pi)\).
Answer
Step 3: However, since the range of \(\sin x\) is \([-1, 1]\), there are no solutions for \(\sin x = -2\). We only need to find the solutions to \(\sin x = 0.5\) which are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\)
Step 1 ：Step 1: We have a quadratic in \(\sin x\), namely \(2\sin^2x - 3\sin x - 2 = 0\). We will factor this into two binomials: \((2\sin x - 1)(\sin x + 2) = 0\)
Step 2 ：Step 2: Setting each factor equal to zero gives us \(2\sin x - 1 = 0\) and \(\sin x + 2 = 0\). Solving these gives us \(\sin x = 0.5\) and \(\sin x = -2\)
Step 3 ：Step 3: However, since the range of \(\sin x\) is \([-1, 1]\), there are no solutions for \(\sin x = -2\). We only need to find the solutions to \(\sin x = 0.5\) which are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\)
