Search any question & find its solution
Question:
Answered & Verified by Expert
The number of solutions to the equation $\cos ^{4} x+\frac{1}{\cos ^{2} x}=\sin ^{4} x+\frac{1}{\sin ^{2} x}$ in the interval $[0,2 \pi]$ is :
Options:
Solution:
1390 Upvotes
Verified Answer
The correct answer is:
4
$\cos ^{4} x-\sin ^{4} x=\frac{1}{\sin ^{2} x}-\frac{1}{\cos ^{2} x}$
$\left(\cos ^{2} x-\sin ^{2} x\right)=\frac{\left(\cos ^{2} x-\sin ^{2} x\right)}{\sin ^{2} x \cos ^{2} x}$
$\cos 2 x=\frac{4 \cos 2 x}{\sin ^{2} 2 x}$
$\cos 2 x\left(1-4 \operatorname{cosec}^{2} 2 x\right)=0$
$\cos 2 x=0$
$2 x=2 n \pi \pm \frac{\pi}{2}$
$x=n \pi \pm \frac{\pi}{4}$
At $n=0, x=\frac{\pi}{4}$
$n=1 ; \quad x=\frac{5 \pi}{4}, \frac{3 \pi}{4}$
$n=2, x=\frac{7 \pi}{4}$
$\left(\cos ^{2} x-\sin ^{2} x\right)=\frac{\left(\cos ^{2} x-\sin ^{2} x\right)}{\sin ^{2} x \cos ^{2} x}$
$\cos 2 x=\frac{4 \cos 2 x}{\sin ^{2} 2 x}$
$\cos 2 x\left(1-4 \operatorname{cosec}^{2} 2 x\right)=0$
$\cos 2 x=0$
$2 x=2 n \pi \pm \frac{\pi}{2}$
$x=n \pi \pm \frac{\pi}{4}$
At $n=0, x=\frac{\pi}{4}$
$n=1 ; \quad x=\frac{5 \pi}{4}, \frac{3 \pi}{4}$
$n=2, x=\frac{7 \pi}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.