Find number of solutions of the equation $\sin^{4} x=1 +\tan^{4} x$ between $0$ and $4\pi.$ \begin{equation*}
\sin^{4} x=1\ +\tan^{4} x \quad \text{ between } \quad  0 \text{ and } 4\pi.
\end{equation*}
I divided LHS and RHS by $\sin^4x$ and tried by substituting $1- \cos(2x)$
in place of $sin^2x$ and $1+cos(2x)$ in place of $cos^2x$ but still I was not able to make out what to do.
 A: Hint: $\sin(x)$ is always between $-1$ and $1$.  
A: Use $\tan x =\frac{\sin x}{\cos x}$ and the identity $\sin2x=2\sin x\cos x$ to express the equation as
$$\sin^42x+8\sin^22x-16=0$$
which has no real solutions.
A: The equation has no solution since $0 \leq (\sin x)^4 \leq 1$ while 
$1+(\tan x)^4 \geq 1$ and for no $x \in \mathbb R$ you have that $\sin x=\pm 1$ and $\tan x=0$.
A: We know that $\tan^4x\ge0$ and $-1\le\sin(x)\le1$ for all $x\in\mathbb{R}$; as a result, we know that $1+\tan^4x\ge1$and $0\le\sin^4x\le1$. As a result, the only possible solution for the equation $$\sin^4x=1+\tan^4x$$ should happen at $\sin^4x=1$, and for $x\in[0,4\pi]$, possible solutions for $\sin^4x=1$ are $x={\pi/2,3\pi/2,5\pi/2,7\pi/2}$. However, if we plug in these $x$'s to the RHS, we found that $\tan^4x$ blows off to infinity, which is obviously greater than one. Accordingly, there are no solutions.
A: Divide both sides by $\sin^4x,$
$$1=\csc^4x+\sec^4x$$
As $\sec^2x,\csc^2x\ge1$
the minimum value of  the right hand side $\ge1+1>1$
