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

The number of solutions of $\cos 2 \theta = \sin \theta$ in the interval $(0, 2 \pi)$ is:

If the sum of solutions of the system of equations $2 \sin^{2} \theta - \cos 2\theta = 0$ and $2 \cos^{2} \theta + 3 \sin \theta = 0$ in the interval $[0, 2\pi]$ is $k\pi$,then $k$ is equal to.

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 $\alpha, \beta, \gamma$ and $\delta$ are the solutions of the equation $\tan \left( \theta + \frac{\pi}{4} \right) = 3 \tan 3\theta$,no two of which have equal tangents,then the value of $\tan \alpha + \tan \beta + \tan \gamma + \tan \delta$ is

Values of $\theta$ $(0^\circ < \theta < 360^\circ)$ satisfying $\text{cosec } \theta + 2 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module