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Question Number 143077 by mathdanisur last updated on 09/Jun/21

$${sin}^{5}x+{cos}^{5}x=2-{sin}^{4}x$$

Answered by MJS_new last updated on 10/Jun/21

$$\sin x=s$$$$\cos x=c$$$${\cos }^{5}x={(1-{\sin }^{2}x)}^2\cos x$$$$\Rightarrow$$$$s^5+{(1-s^2)}^{2}c=2-s^4$$$$s^5+(c+1)s^4-2{cs}^2+c-2=0$$$$(s-1)((s^3+s^2-s-1)c+(s^4+2s^3+2s^2+2s+2))=0$$$$s_1=1\Rightarrow \sin x=1\Rightarrow \bullet x=\frac{\pi }{2}+2n\pi \bullet$$$$(s^3+s^2-s-1)c+(s^4+2s^3+2s^2+2s+2)=0$$$$c=f\left(s\right)=\frac{s^4+2s^3+2s^2+2s+2}{(1-s){(1+s)}^2}$$$$-1⩽s⩽1\Rightarrow f\left(s\right)⩾2\Rightarrow \mathrm{no}\mathrm{other}\mathrm{solution}$$

Commented by mathdanisur last updated on 10/Jun/21

$$nicethankssir$$

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