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

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(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

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

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(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

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