The number of solutions of the equation cos^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equation: $\cos^2 2x + \cos^2 \frac{5x}{4} = \cos 2x \cos^2 5x$.For the equation to hold,we analyze the range of the functions.Note that $\c

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given equation: $\cos^2 2x + \cos^2 \frac{5x}{4} = \cos 2x \cos^2 5x$.For the equation to hold,we analyze the range of the functions.Note that $\cos^2 2x - \cos 2x \cos^2 5x + \cos^2 \frac{5x}{4} = 0$.This is a quadratic in $\cos 2x$: $(\cos 2x)^2 - (\cos^2 5x)\cos 2x + \cos^2 \frac{5x}{4} = 0$.For real solutions,the discriminant $D \ge 0$:$D = (\cos^2 5x)^2 - 4 \cos^2 \frac{5x}{4} \ge 0$.Since $\cos^2 5x \le 1$ and $\cos^2 \frac{5x}{4} \ge 0$,the only way for this to hold is if $\cos 2x = 0$ and $\cos \frac{5x}{4} = 0$.Solving $\cos 2x = 0$ in $[0, \frac{\pi}{3}]$ gives $2x = \frac{\pi}{2} \implies x = \frac{\pi}{4}$.Checking $x = \frac{\pi}{4}$ in $\cos \frac{5x}{4} = 0$: $\cos \frac{5\pi}{16} 
eq 0$.Thus,there are no solutions in the given interval.

The number of solutions of the equation $\cos^2 2x + \cos^2 \frac{5x}{4} = \cos 2x \cos^2 5x$ in the interval $[0, \frac{\pi}{3}]$ is:

Similar Questions

The number of values of $x$ satisfying $\sin 4x = \cos 3x$ in the interval $\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ is:

If $\sec 4\theta - \sec 2\theta = 2$,then the general value of $\theta$ is

If $\tan 3 \theta = \cot \theta$,then $\theta = $

The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is

The equation that is satisfied by the general solution of the equation $4-3 \cos ^2 \theta=5 \sin \theta \cos \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module