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 $x=0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}$.
We don't need any factorization, indeed since for $x\neq k\frac \pi 2$ we have that $0<|\sin x|<1$ and $0<|\cos x|<1$ then $$\sin^8 {x}+\cos^6 {x}<\sin^2 {x}+\cos^2 {x}=1$$
we need only to check for the solutions $x= k\frac \pi 2$ which indeed are all the solutions.
Hint: Write your equation in the form $$- \left( \sin \left( x \right) \right) ^{2} \left( \cos \left( x \right) \right) ^{2} \left( \left( \cos \left( x \right) \right) ^ {4}-3\, \left( \cos \left( x \right) \right) ^{2}+4 \right) =0$$
Let $\cos x=u\implies \sin x=\sqrt{1-u^2}$. Then $$\sin^8x+\cos^6x=(1-u^2)^4+u^6=(1-4u^2+6u^4-4u^6+u^8)+u^6=1$$ gives $$u^8-3u^6+6u^4-4u^2=0\implies u^2(u^2-1)(u^4-2u^2+4)=0$$ so $u=\cos x=0,\pm1$ as $\Delta_{v^2-2v+4}<0$.
