<s>[INST] 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 ?
 [/INST] 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 $\Delta_{v^2-2v+4}<0$.
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