Q43 / 90 · KVPY 2014 (SB/SX) · Mathematics

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 :

  • A
    6
  • Correct answer
    B
    4
  • C
    2
  • D
    0

Solution

$\cos ^{4} x-\sin ^{4} x=\frac{1}{\sin ^{2} x}-\frac{1}{\cos ^{2} x}$ $\left(\cos ^{2} x-\sin ^{2} x\right)=\frac{\left(\cos ^{2} x-\sin ^{2} x\right)}{\sin ^{2} x \cos ^{2} x}$ $\cos 2 x=\frac{4 \cos 2 x}{\sin ^{2} 2 x}$ $\cos 2 x\left(1-4 \operatorname{cosec}^{2} 2 x\right)=0$ $\cos 2 x=0$ $2 x=2 n \pi \pm \frac{\pi}{2}$ $x=n \pi \pm \frac{\pi}{4}$ At $n=0, x=\frac{\pi}{4}$ $n=1 ; \quad x=\frac{5 \pi}{4}, \frac{3 \pi}{4}$ $n=2, x=\frac{7 \pi}{4}$

Hence, the correct answer is option: 'B'.

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