Q. The solution of equation $\cos^{2}{\theta + \sin{\theta + 1 = 0}}$ lies in the interval
A) $\left( - \frac{\pi}{4},\ \frac{\pi}{4} \right)$
B) $\left( \frac{\pi}{4},\ \frac{3\pi}{4} \right)$
C) $\left( \frac{3\pi}{4},\ \frac{5\pi}{4} \right)$
D) $\left( \frac{5\pi}{4},\ \frac{7\pi}{4} \right)$
Explanation
(d)
Given, $\cos^{2}{\theta + \sin^{2}{\theta + 1 = 0}}$
$\Rightarrow \ \sin^{2}\theta - \sin{\theta - 2 = 0}$
$\Rightarrow \ \ \left( \sin{\theta + 1} \right)\left( \sin{\theta - 2} \right) = 0$
$\Rightarrow \ \ \sin{\theta = - 1 = \sin{\frac{3\pi}{2}\ \ \ \ \ \ \ \lbrack\because\sin{\theta \geq 1}\rbrack}}$
$\therefore\ \ \ \theta = \frac{3\pi}{2} \in \left( \frac{5\pi}{4},\frac{7\pi}{4} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.