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Q.
The number of solutions of equation $\sin^4\, \theta-2 \sin^2\, \theta + 1 = 0$ which lie between $0$ and $2\pi$ is
Trigonometric Functions
Solution:
From the given equation, we have
$\sin^2 \, \theta = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$
Now $\sin^2 \, \theta \neq 1 + \sqrt{2} \,$
$ \because \, 1 + \sqrt{2} > 1$
Also $\sin^2 \, \theta \neq 1 - \sqrt{2}$
$\therefore $ $1 - \sqrt{2} < 0 $ and $\sin^2 \, \theta$ cannot be negative
$\therefore $ no value of $\theta$ satisfy the equation, hence no root lies in 0 and $2 \pi$.