The real roots of the equation cos^7x + sin^4 | vedclass

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

Question

(B) Given equation: $\cos^7x + \sin^4x = 1$We know $\sin^4x = (1 - \cos^2x)^2 = 1 - 2\cos^2x + \cos^4x$.Substituting this into the equation: $\cos^7x

Vedclass · Accepted Answer

$\{ -\frac{\pi}{2}, 0, \frac{\pi}{2} \}$

Vedclass · Answer

(B) Given equation: $\cos^7x + \sin^4x = 1$We know $\sin^4x = (1 - \cos^2x)^2 = 1 - 2\cos^2x + \cos^4x$.Substituting this into the equation: $\cos^7x + 1 - 2\cos^2x + \cos^4x = 1$$\cos^7x + \cos^4x - 2\cos^2x = 0$$\cos^2x(\cos^5x + \cos^2x - 2) = 0$Case $1$: $\cos^2x = 0 \implies \cos x = 0 \implies x = \pm \frac{\pi}{2}$.Case $2$: $\cos^5x + \cos^2x - 2 = 0$.Let $f(t) = t^5 + t^2 - 2$ where $t = \cos x$. Since $t \in [-1, 1]$,the maximum value of $t^5 + t^2$ is $1^5 + 1^2 = 2$.Thus,$t^5 + t^2 - 2 = 0$ only when $t = 1$,which means $\cos x = 1$.$\cos x = 1 \implies x = 0$.Combining the results,the roots in $(-\pi, \pi)$ are $\{ -\frac{\pi}{2}, 0, \frac{\pi}{2} \}$.

The real roots of the equation $\cos^7x + \sin^4x = 1$ in the interval $(-\pi, \pi)$ are

Similar Questions

If $\sin \theta + \cos \theta = 1$,then the general value of $\theta$ is

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

The total number of solutions of $\sin^4x + \cos^4x = \sin x \cos x$ in the interval $[0, 2\pi]$ is equal to

The general solution of $\frac{\tan 2x - \tan x}{1 + \tan x \tan 2x} = 1$ is:

The number of solutions of the equation $2 \theta - \cos^{2} \theta + \sqrt{2} = 0$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module