Mathematics | VITEEE 2016 | Discussion Page

The solution of the equation $\cos^{2}\theta+\sin^{2}\theta+1=0$ , lies in the interval

$\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$

$\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$

$\left(\frac{3\pi}{4},\frac{5\pi}{4}\right)$

$\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)$

Option D

We have ,

$\cos^{2}\theta+\sin^{2}\theta+1=0\Rightarrow 1-\sin^{2}+\sin\theta+1=0$

$\Rightarrow \sin\theta=-1(\because\sin\theta\neq2)\Rightarrow \theta=3\pi/2$

$\therefore$ $\theta\in \left(\frac{5\pi}{4},\frac{7\pi}{4}\right)$

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