The number of solutions of sin^{7} x + cos^{7 | vedclass

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(A) Given $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$.Since $\sin^{2} x \leq 1$ and $\cos^{2} x \leq 1$,for $x \in [0, \pi/2]$,we have $\sin^{

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$5$

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(A) Given $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$.Since $\sin^{2} x \leq 1$ and $\cos^{2} x \leq 1$,for $x \in [0, \pi/2]$,we have $\sin^{7} x \leq \sin^{2} x$ and $\cos^{7} x \leq \cos^{2} x$.Adding these,$\sin^{7} x + \cos^{7} x \leq \sin^{2} x + \cos^{2} x = 1$.The equality holds only when $\sin^{7} x = \sin^{2} x$ and $\cos^{7} x = \cos^{2} x$.This implies $(\sin x = 0 	ext{ or } \sin x = 1)$ and $(\cos x = 0 	ext{ or } \cos x = 1)$.Case $1$: $\sin x = 0 \implies x = 0, \pi, 2\pi, 3\pi, 4\pi$. Checking $\cos^{7} x = 1$,we get $x = 0, 2\pi, 4\pi$.Case $2$: $\cos x = 0 \implies x = \pi/2, 3\pi/2, 5\pi/2, 7\pi/2$. Checking $\sin^{7} x = 1$,we get $x = \pi/2, 5\pi/2$.Combining these,the solutions are $x \in \{0, \pi/2, 2\pi, 5\pi/2, 4\pi\}$.Thus,there are $5$ solutions.

The number of solutions of $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$ is equal to:

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