Question

The number of real solutions in the interval \(\left[-\frac{3\pi}{2}, \frac{7\pi}{2}\right]\) of the equation \(4\sin^{2}x-8\sin x+3=0\) is ________.

Hint:

Factorise as a quadratic in \(\sin x\); reject \(\sin x=\frac32\) and count the solutions of \(\sin x=\frac12\) in the given interval.

Answer 1

Correct Answer: 5

\(4\sin^2x-8\sin x+3=0\Rightarrow(2\sin x-1)(2\sin x-3)=0\), so \(\sin x=\frac12\) or \(\frac32\).
\(\sin x=\frac32\) is impossible. So \(\sin x=\frac12\), i.e. \(x=\frac{\pi}{6}+2k\pi\) or \(x=\frac{5\pi}{6}+2k\pi\).
Interval: \(-\frac{3\pi}{2}\le x\le\frac{7\pi}{2}\).
Family \(\frac{\pi}{6}+2k\pi\): \(k=0,1\) give \(\frac{\pi}{6},\frac{13\pi}{6}\) (k=-1 gives \(-\frac{11\pi}{6}\), below \(-\frac{3\pi}{2}\); k=2 gives \(\frac{25\pi}{6}\), above \(\frac{7\pi}{2}\)): 2 solutions.
Family \(\frac{5\pi}{6}+2k\pi\): \(k=-1,0,1\) give \(-\frac{7\pi}{6},\frac{5\pi}{6},\frac{17\pi}{6}\) (k=2 gives \(\frac{29\pi}{6}\), above \(\frac{7\pi}{2}\)): 3 solutions.
Total \(=2+3=5\).
Answer: 5