KVPY SX SB Exam 20141 Question Paper

Question : 8 of 120

Marks: +1, -0

The number of solutions to the equation

{\cos^4x}+1/{\cos^2x}={\sin^4x}+1/{\sin^2x}

in the interval [0 ,2π] is

Solution:

{\cos^4x}-{\sin^4x}=1/{\sin^2x}-1/{\cos^2x}

(\cos^2 x - \sin^2 x) = (\cos^2 x - \sin^2 x) /{\sin^2x\cos^2x}

\cos 2x ={4\cos 2x}/{\sin ^2 2x}

\cos 2x (1- 4 \cosec^22x) = 0

cos 2x = 0

2x = 2nπ ±π/2

x = nπ ±π/4

\At n = 0, x = π/4

n = 1; x = {5π}/4,{3π}/4

n = 2, x = {7π}/4

Go to Question:

Prev Question

Next Question