The general solution of the equation sin^{100 | vedclass

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

Question

(B) Given the equation $\sin^{100} x - \cos^{100} x = 1$.This can be rewritten as $\sin^{100} x = 1 + \cos^{100} x$.We know that for all real $x$,$\si

Vedclass · Accepted Answer

$n\pi + \frac{\pi}{2}, n \in I$

Vedclass · Answer

(B) Given the equation $\sin^{100} x - \cos^{100} x = 1$.This can be rewritten as $\sin^{100} x = 1 + \cos^{100} x$.We know that for all real $x$,$\sin^{100} x \le 1$ and $\cos^{100} x \ge 0$,which implies $1 + \cos^{100} x \ge 1$.For the equality to hold,we must have $\sin^{100} x = 1$ and $1 + \cos^{100} x = 1$.From $1 + \cos^{100} x = 1$,we get $\cos^{100} x = 0$,which implies $\cos x = 0$.This occurs when $x = n\pi + \frac{\pi}{2}$ for $n \in I$.Checking these values in $\sin^{100} x = 1$,we have $\sin(n\pi + \frac{\pi}{2}) = \pm 1$,so $\sin^{100}(n\pi + \frac{\pi}{2}) = (\pm 1)^{100} = 1$.Thus,the general solution is $x = n\pi + \frac{\pi}{2}, n \in I$.

The general solution of the equation $\sin^{100} x - \cos^{100} x = 1$ is

Similar Questions

Find the general solution of $\sin x = -\frac{\sqrt{3}}{2}$.

The principal solutions of $\tan 3 \theta = -1$ are

The general solution of $\tan 3x = 1$ is

If $\sin^2 x + \sin x \cos x - 6\cos^2 x = 0$ and $-\frac{\pi}{2} < x < 0$,then the value of $\cos 2x$ is

The number of real values of $x \in [0, 2\pi] - \{\frac{\pi}{2}, \frac{3\pi}{2}\}$ satisfying the equation $|\cos x|^{2\sin^2 x - 3\sin x + 1} = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module