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sum of all the real values of x satisfying sin^3 x + cos^5 x=1 on [0,4*pie] is equal to k* pie for some integer k where k is equal to x. find x.

sum of all the real values of x satisfying sin^3 x + cos^5 x=1 on [0,4*pie] is equal to k* pie for some integer k where k is equal to x. find x.
 

Grade:12

2 Answers

jagdish singh singh
173 Points
5 years ago
\hspace{-0.6 cm}$Given $\sin^3 x+\cos^5 x=1\;,$ Using $\sin^3 x\leq \sin^2 x$ and $\cos^5 x\leq \cos^2 x$\\\\So we get $\sin^3 x+\cos^5 x\leq \sin^2 x+\cos^2 x=1$\\\\So here equality hold when $\sin^3 x=\sin^2 x$ and $\cos^5 x= \cos^2 x$\\\\
\hspace{-0.6 cm}$So we have two cases arise\\\\$\bullet\; \sin^2 x=1\Rightarrow x=n\pi\pm \frac{\pi}{2}$ and $\cos^2 x=0\Rightarrow x=n\pi\pm \frac{\pi}{2}$\\\\$\bullet\; \sin^2 x=0\Rightarrow x=n\pi$ and $\cos^2 x=1\Rightarrow x=n\pi$\\\\So here Solution Set of $x\in \left[0,4\pi\right]$\\\\ So we get $x=\frac{\pi}{2}\;,\frac{3\pi}{2}\;,\frac{5\pi}{2}\;,\frac{7\pi}{2}$ and $x=0\;,\pi\;,2\pi\;,3\pi,4\pi.$
 
\hspace{-0.6 cm}$So Sum of all Roots is $=\frac{\pi}{2}+\frac{3\pi}{2}+\frac{5\pi}{2}+\frac{7\pi}{2}+0+\pi+2\pi+3\pi+4\pi$\\\\So Sum of real roots is $ = 8\pi+10\pi=18\pi.$
jagdish singh singh
173 Points
5 years ago
Above answer is Wrong So plz dont consider it. Thanks........................
 
I have calculate it wrongly.

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