The solution of the equation cos^2 theta + si | vedclass

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Question

(D) Given equation: $\cos^2 	heta + \sin 	heta + 1 = 0$Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get:$1 - \sin^2 	heta + \sin 	het

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

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(D) Given equation: $\cos^2 	heta + \sin 	heta + 1 = 0$Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we get:$1 - \sin^2 	heta + \sin 	heta + 1 = 0$$-\sin^2 	heta + \sin 	heta + 2 = 0$Multiplying by $-1$:$\sin^2 	heta - \sin 	heta - 2 = 0$Factoring the quadratic equation:$(\sin 	heta - 2)(\sin 	heta + 1) = 0$This gives $\sin 	heta = 2$ or $\sin 	heta = -1$.Since the range of $\sin 	heta$ is $[-1, 1]$,$\sin 	heta = 2$ is impossible.Thus,$\sin 	heta = -1$.The general solution is $	heta = 2n\pi - \frac{\pi}{2}$.For $n=1$,$	heta = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.Since $\frac{5\pi}{4} < \frac{3\pi}{2} < \frac{7\pi}{4}$,the solution lies in the interval $\left( \frac{5\pi}{4}, \frac{7\pi}{4} ight)$.

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

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