In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2 {x})(1+\sin^4 {x}) , \cos^2{x}=0$
$(1-\sin^2{x})^2=(1+\sin^2 {x})(1+\sin^4 {x})$
$-3\sin^2{x}=\sin^6{x}$
Which is not possilbe
Is there a trick or something to solve this equation or to know how many solutions are there ?

One trick is letting $t=\sin^2(x)$, so that the equation reduces to a polynomial $t^4+(1-t)^3 - 1 = t^4 - t^3 + 3t^2 - 3t = t(t^3-t^2+3t-3) = t(t-1)(t^2+3)=0$
Thus either $t=0,1$ so that $\sin(x) = 0,1,-1$. In $[0, 2\pi)$, we clearly have solutions $x=0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}$. 
