chota pack ??

Since sin⁴x ≤ sin²x and cos¹⁵x ≤ cos²x. Hence
\sin^4x + \cos^{15}x\leq \sin^2x +\cos^2x=1
So that we always have
\sin^4x + \cos^{15}x\leq 1
So according to the given condition, the equality in the above inequality holds. So we must have
\sin^4x =\sin^2x
and
\cos^{15}x=\cos^2x
So we get \sin x = \{0,-1,1\}
And \cos x=\{1,0\}
So x is either an even multiple of \pi or an odd multiple of \pi/2.