Period of $\sin^{6}(x)+\cos^{4}(x)$ I am trying to find the fundamental period of $\sin^{6}(x)+\cos^{4}(x)$. Now the period of $\sin^{6}(x)$ is $\pi$ and that of $\cos^{4}(x)$ is also $\pi$, so expectedly the period of the expression should be the $\text{LCM}\{{\pi,\pi}\}=\pi$. But obviously, since they are even, the period can be less than the LCM obtained.

Update 1:
It can be used that expression will repeat between two $x$-values where derivative is $0$. So solving that gives one of the values as $x_0=\arcsin\left(\sqrt{\frac{-1+\sqrt{7}}{3}}\right)$, and so period should be $2x_0$ and that is confirmed with graph, but how to go about it in a more intuitive way? 

I am not able to come up with methods to solve the period of this expression. Any hints are appreciated. Thanks. 
 A: Let $f(x)=\sin^6x+\cos^4x$ and $T>0$ be a minimal period.
Thus, $$f(0+T)=f(0)$$ or $$\sin^6T+\cos^4T=1$$ or
$$\sin^2T(\sin^4T-1-\cos^2T)=0$$ or
$$\sin^2T\cos^2T(2+\sin^2T)=0,$$ which gives $$T\geq\frac{\pi}{2}$$ because for any $0<T<\frac{\pi}{2}$ the equality $$\sin^2T\cos^2T(2+\sin^2T)=0$$ is wrong.
But easy to see that $$f\left(x+\frac{\pi}{2}\right)\neq f(x).$$
Thus, $$T\geq\pi$$ and easy to check that $\pi$ is indeed a period.
A: We have to compute the first derivate:
$$f'(x)=2\sin(x)\cos(x)(3\sin^4(x)+2\sin^2(x)-2)$$
To find the maxima and minima point, we have to solve:
$$\sin(2x)(3\sin^4(x)+2\sin^2(x)-2)=0$$
We can break it down in:
$$\sin(2x)=0 \leftrightarrow x=0+k\frac{\pi}{2}\;\;\;\;\; (1)$$
And
$$3\sin^4(x)+2\sin^2(x)-2=0\rightarrow t=\sin^2(x) \;\;t\in R^+ \rightarrow 3t^2+2t-2=0 \leftrightarrow t=\frac{-2\pm\sqrt{28}}{6}=\frac{-1+\sqrt{7}}{3},\frac{-1-\sqrt{7}}{3} \;\;\;\text{IMPOSSIBLE}$$
From this, we have: $$x=\arcsin(\frac{-1+\sqrt{7}}{3})+2k\pi \vee x=\pi-\arcsin(\frac{-1+\sqrt{7}}{3})+2k\pi$$
Now, the period is the distance between two maximum points that are given by $(1)$, so:
$$T=\frac{\pi}{2}-0=\frac{\pi}{2}$$
