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Trigonometrical Equations
normal
Number of solutions of $\sqrt {\tan \theta } = 2\sin \theta ,\theta \in \left[ {0,2\pi } \right]$ is equal to
A
$2$
B
$4$
C
$5$
D
$6$
Solution
Let $\tan \theta=t \Rightarrow \theta \in[0, \pi / 2] \cup\{\pi, 2 \pi\}$
$\Rightarrow \sqrt{t}=2 \sin \theta$
squaring
$\left(1+\frac{1}{t^{2}}\right) t=4$
$\mathrm{t}+\frac{1}{\mathrm{t}}=4 \Rightarrow \mathrm{t}=2+\sqrt{3}, 2-\sqrt{3}$
$\Rightarrow \theta=0, \frac{\pi}{12}, \frac{5 \pi}{12}, \pi, 2 \pi$
Standard 11
Mathematics
