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The solution of $(2 \cos x-1)(3+2 \cos x)=0$ in the interval $0 \leq x \leq 2 \pi$ is
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The correct answer is:
$\frac{\pi}{3}, \frac{5 \pi}{3}$
We have $(2 \cos x-1)(3+2 \cos x)=0$
If $2 \cos x-1=0,$ then $\cos x=\frac{1}{2}$
$\therefore x=\pi / 3,5 \pi / 3$
If $3+2 \cos x=0,$ the $\cos x=-3 / 2$
which is not possible.
If $2 \cos x-1=0,$ then $\cos x=\frac{1}{2}$
$\therefore x=\pi / 3,5 \pi / 3$
If $3+2 \cos x=0,$ the $\cos x=-3 / 2$
which is not possible.
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