# 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.

jagdish singh singh
173 Points
8 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
8 years ago
Above answer is Wrong So plz dont consider it. Thanks........................

I have calculate it wrongly.