let $k$ be a parameter then equation $\cos x=k \sin (\frac{x}{4})$ how many answers in $(0,8\pi)$? let $k$ be  a parameter  then equation $\cos x=k \sin (\frac{x}{4})$ how many  answers in $(0,8\pi)$? 
My Try : $x=4t$ then $\cos 4t=k\sin t$ 
now we have :
$$\sin^4 t+\cos ^4 t-6\sin^2 t \cos ^2t -k\sin t=0$$
now  what do I do ?
 A: We have
$$\cos x = 1 - 2\sin^2 \frac{x}{2} = 1 - 8\sin^2 \frac{x}{4}\cos^2 \frac{x}{4}$$
Let $t = \sin\frac{x}{4}$, where $t\in [-1,1]$, then the equation becomes
$$ 1 - 8t^2(1-t^2) = kt $$
$$ 8t^4 - 8t^2 + 1 = kt $$
Since $\sin \frac{x}{4}$ has a period of $8\pi$, we only need to determine the number of solutions in $t$, which happens to be the intersection of two functions
$$ \left\{ \begin{aligned} f(t) &= 8t^4 - 8t^2 + 1 \\ g(t) &= kt \end{aligned} \right. $$
Here is a graph with different values of $k$. The graph of $f(t)$ is in red. 

Due to symmetry, $-k$ and $k$ give the same number of solutions, so we only need to consider the range of $|k|$.
There is a special value of $|k|$ for which $kt$ is exactly tangent to $f(t)$ (approximately $\sim 1.47$). 


*

*For $|k| \le 1$ (orange line), there are 4 solutions. 

*For $1 < |k| < 1.47$ (purple line), there are 3 solutions.

*For $|k| = 1.47$ (green line), there are 2 solutions.

*For $|k| > 1.47$ (black line), there is one solution.



Note: The exact value of $|k|$ where the two graphs are tangent is a solution of the system
$$ \left\{ \begin{aligned} 8t^4 - 8t^2 + 1 &= kt \\ 32t^3 - 16t &= k \end{aligned} \right. $$
I don't believe there is a closed form. Here is a numerical solution given by WolframAlpha.

Note 2: Each value of $t$ corresponds to two values of $x$, except when $t = \pm 1$ (which occurs when $|k|=1$), so the number of solutions in $x$ breaks down further


*

*For $|k| < 1$, there are 8 solutions.

*For $|k| = 1$, there are 7 solutions. 

*For $1 < |k| < 1.47$, there are 6 solutions.

*For $|k| = 1.47$, there are 4 solutions.

*For $|k| > 1.47$, there are 2 solutions.


Just for fun, here is another graph in $x$

