Find all real solution for k: $2\sin^2(x)+6\cos^2\left(\frac x4\right)=5-2k$ I have been stuck at this problem for some time now. I'd really apprechiate your help. Thanks.
$$2\sin^2(x)+6\cos^2(\frac x4)=5-2k$$
 A: Too long for a comment but I cannot resist when I see an equation !
As @lhf answered, we are looking for the maximum of
$$f(x)=2 \sin ^2(x)+6 \cos ^2\left(\frac{x}{4}\right)$$ so for the zero's of
$$f'(x)=2 \sin (2 x)-\frac{3}{2} \sin \left(\frac{x}{2}\right)$$
Let $x=2 \sin ^{-1}(t)$ to get as a new equation
$$8 t \left(1-2 t^2\right) \sqrt{1-t^2}-\frac{3 t}{2}=0$$ Discarding the trivial $t=0$ and squaring
$$1024 t^6-2048 t^4+1280 t^2-247=0$$ which is a cubic in $t^2$ and, using $z=t^2$, the solutions are given by
$$z_k=\frac{2}{3}+\frac{1}{3} \cos \left(\frac{2 \pi  k}{3}-\frac{1}{3} \cos
   ^{-1}\left(\frac{13}{256}\right)\right) \qquad \text{with} \qquad k=0,1,2$$ Back to $t$ and $x$, the final solution is
$$x_*=2 \sin ^{-1}\left(\sqrt{\frac{2}{3}-\frac{1}{3} \sin \left(\frac{\pi }{6}+\frac{1}{3} \cos^{-1}\left(\frac{13}{256}\right)\right)}\right)\approx 1.33019$$ and then
$$f(x_*)=4+\sqrt{3\left(1+\cos \left(\frac{\pi }{6}+\frac{1}{3} \sin ^{-1}\left(\frac{13}{256}\right)\right) \right) }+$$ $$\cos \left(2 \sin ^{-1}\left(-\frac 13+\frac 23\sin \left(\frac{\pi }{6}+\frac{1}{3} \cos ^{-1}\left(\frac{13}{256}\right)\right)\right)\right)\approx 7.24701$$
Edit
This is exactly the same result as the elegant one provided by @bjorn93 (which, I must confess, I did  not pay attention when I started working the problem).
A: HINT
If you are trying to solve for $k$ in terms of $x$, why not use the fact that $\sin^2 a + \cos^2 a = 1$ for all real $a$, and eliminate $\sin^2$ on the LHS, and then use $\arccos$ to do what you need?
A: Hint: The problem is about finding the minimum and maximum values of $
2\sin^2(x)+6\cos^2\left(\frac x4\right)$.
The minimum value is easy: it's $0$, taken at $x=10\pi$ for instance.
The maximum value is not easy: it's approximately $M=7.24701$.
This gives $0 \le 5-2k \le M$, or $(5 - M)/2 \le k \le 5/2$.
