Question
If $$4\,{\sin ^2}x - 8\sin x + 3 \leqslant 0,0 \leqslant x \leqslant 2\pi ,$$ then the solution set for $$x$$ is
A.
$$\left[ {0,\frac{\pi }{6}} \right]$$
B.
$$\left[ {0,\frac{5\pi }{6}} \right]$$
C.
$$\left[ {\frac{{5\pi }}{6},2\pi } \right]$$
D.
$$\left[ {\frac{\pi }{6},\frac{{5\pi }}{6}} \right]$$
Answer :
$$\left[ {\frac{\pi }{6},\frac{{5\pi }}{6}} \right]$$
Solution :
Here, $$\left( {2\sin x - 1} \right)\left( {2\sin x - 3} \right) \leqslant 0.$$ But $$2\sin x - 3$$ is always negative.
$$\therefore \,\,2\sin x - 1 \geqslant 0,\,{\text{i}}{\text{.e}}{\text{.,}}\sin x \geqslant \frac{1}{2}.$$
∴ from the figure, $$\frac{\pi }{6} \leqslant x \leqslant \frac{{5\pi }}{6}.$$
Here, $$\left( {2\sin x - 1} \right)\left( {2\sin x - 3} \right) \leqslant 0.$$ But $$2\sin x - 3$$ is always negative.
$$\therefore \,\,2\sin x - 1 \geqslant 0,\,{\text{i}}{\text{.e}}{\text{.,}}\sin x \geqslant \frac{1}{2}.$$
∴ from the figure, $$\frac{\pi }{6} \leqslant x \leqslant \frac{{5\pi }}{6}.$$