The number of solutions to the equation cos^4 | vedclass

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

Question

(B) Given equation: $\cos^4 x + \frac{1}{\cos^2 x} = \sin^4 x + \frac{1}{\sin^2 x}$Rearranging the terms: $\cos^4 x - \sin^4 x = \frac{1}{\sin^2 x} -

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given equation: $\cos^4 x + \frac{1}{\cos^2 x} = \sin^4 x + \frac{1}{\sin^2 x}$Rearranging the terms: $\cos^4 x - \sin^4 x = \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x}$Using the identity $a^2 - b^2 = (a-b)(a+b)$: $(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}$Since $\cos^2 x + \sin^2 x = 1$,we have: $\cos^2 x - \sin^2 x = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x}$This implies $(\cos^2 x - \sin^2 x) \left(1 - \frac{1}{\sin^2 x \cos^2 x}\right) = 0$Using $\sin 2x = 2 \sin x \cos x$,we get $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$.So,$\cos 2x \left(1 - \frac{4}{\sin^2 2x}\right) = 0$This gives $\cos 2x = 0$ or $\sin^2 2x = 4$. Since $\sin^2 2x$ cannot be $4$,we must have $\cos 2x = 0$.Thus,$2x = (2n+1)\frac{\pi}{2}$,which means $x = (2n+1)\frac{\pi}{4}$.For $x \in [0, 2\pi]$,the solutions are $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.Therefore,the total number of solutions is $4$.

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

Similar Questions

If $\sin^2 \theta = \frac{1}{4},$ then the most general value of $\theta$ is

If $0 \leq x \leq 2 \pi$,then the number of real values of $x$ which satisfy the equation $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$ is

If the general solution of $\sin x + 3 \sin 3x + \sin 5x = 0$ is $x = y$,then the set of all values of $\cos y$ is

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

The general solution of $\cos 4\theta = \cos 3\theta$ is $\theta =.....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module