Question

If $3si{n}^{2}x-8sinx+3=0,0\le x\le 2\pi$, then the solution set for x is

• A. $[0,\frac{\pi }{6}]$
• B. $[0,\frac{5\pi }{6}]$
• C. $[\frac{5\pi }{6},2\pi ]$
• D. $[\frac{\pi }{6},\frac{5\pi }{6}]$

Answer 1

Answer: (D)

$[\frac{\pi }{6},\frac{5\pi }{6}]$

$3{\sin }^{2}x-8\sin x+3=0$

$\therefore \sin x=\frac{8\pm \sqrt{64-36}}{6}=\frac{8\pm \sqrt{28}}{6}=\frac{8\pm 2\sqrt{7}}{6}=\frac{4\pm \sqrt{7}}{3}$

$\sin x=\frac{4+\sqrt{7}}{3}>1$ or $\sin x=\frac{4-\sqrt{7}}{3}$ occurs twice in $[0,2\pi ]$ range.