Question
If $$2\sin x + 1 \geqslant 0$$ and $$x \in \left[ {0,2\pi } \right]$$ then the solution set for $$x$$ is
A.
$$\left[ {0,\frac{{7\pi }}{6}} \right]$$
B.
$$\left[ {0,\frac{{7\pi }}{6}} \right] \cup \left[ {\frac{{11\pi }}{6},2\pi } \right]$$
C.
$$\left[ {\frac{{11\pi }}{6},2\pi } \right]$$
D.
None of these
Answer :
$$\left[ {0,\frac{{7\pi }}{6}} \right] \cup \left[ {\frac{{11\pi }}{6},2\pi } \right]$$
Solution :
$$2\sin x + 1 \geqslant 0$$
$$ \Rightarrow \,\,\sin x \geqslant - \frac{1}{2}.$$ The value scheme for this is shown below.
From the figure,
$$\eqalign{ & \frac{{11\pi }}{6} \leqslant x \leqslant 2\pi \,\,{\text{or, }}0 \leqslant x \leqslant \frac{{7\pi }}{6}. \cr & \therefore \,\,x \in \left[ {0,\frac{{7\pi }}{6}} \right] \cup \left[ {\frac{{11\pi }}{6},2\pi } \right]. \cr} $$
$$2\sin x + 1 \geqslant 0$$
$$ \Rightarrow \,\,\sin x \geqslant - \frac{1}{2}.$$ The value scheme for this is shown below.
From the figure,
$$\eqalign{ & \frac{{11\pi }}{6} \leqslant x \leqslant 2\pi \,\,{\text{or, }}0 \leqslant x \leqslant \frac{{7\pi }}{6}. \cr & \therefore \,\,x \in \left[ {0,\frac{{7\pi }}{6}} \right] \cup \left[ {\frac{{11\pi }}{6},2\pi } \right]. \cr} $$