Skip to main content
Back to Trigonometric Equations
JEE Main 2018
Trigonometric Equations
Trigonometric Equations
Hard

Question

The number of solutions of the equation 1 + sin 4 x = cos 2 3x, x[5π2,5π2]x \in \left[ { - {{5\pi } \over 2},{{5\pi } \over 2}} \right] is :

Options

Solution

1+sin4x1=cos23x1\mathop {\underline {1 + {{\sin }^4}x} }\limits_{ \ge 1} = \mathop {\underline {{{\cos }^2}3x} }\limits_{ \le 1} Now for equality to hold sin 4 x = 0 & cos 2 3x = 1 sin 4 = 0 \Rightarrow x = -2π\pi , - π\pi , 0, π\pi , 2π\pi All of which satisfy cos 2 3x = 1 \Rightarrow 5 solutions

Practice More Trigonometric Equations Questions

View All Questions