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Q. The number of principal solutions of $\tan 2 \theta = 1$ is

MHT CETMHT CET 2017Trigonometric Functions

Solution:

Let $y=\tan 2 \theta$
$\tan 2 \theta=1$
$\Longrightarrow 2 \theta=\tan ^{-1}(1)$
$\Longrightarrow 2 \theta=\frac{\pi}{4}+ k \pi$, where $k$ is a positive integer
Now,
for $k =0 \Rightarrow \theta=\frac{\pi}{4}$ and for $k = 1 \Rightarrow \theta =\frac{ 5 \pi}{4}$
$0 \leq \frac{\pi}{4} \leq 2 \pi$ and $0 \leq \frac{5 \pi}{4} \leq 2 \pi$
$\therefore 2 \theta=\frac{\pi}{4}, \frac{5 \pi}{4}$
$\Longrightarrow \theta=\frac{\pi}{8}, \frac{5 \pi}{8}$
Hence, the required principal solutions are $\frac{\pi}{8}$ and $\frac{5 \pi}{8}$