(D) The given equation is $\tan 2\theta = 1$. Since $\tan 2\theta$ is positive,$2\theta$ lies in the first or third quadrant. For the principal soluti

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(D) The given equation is $\tan 2\theta = 1$. Since $\tan 2\theta$ is positive,$2\theta$ lies in the first or third quadrant. For the principal solutions,we consider $0 \le \theta Thus,$2\theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$. Dividing by $2$,we get $\theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$. Hence,there are $4$ principal solutions.

Class 11 Mathematics Trigonometrical Equations Solution of trigonometrical equations

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

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