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Question Number 140907 by bramlexs22 last updated on 14/May/21

(√(sin x)) cos x −(√(sin^5 x)) cos x = cos^3 x (√(sin x))

$$\:\sqrt{\mathrm{sin}\:\mathrm{x}}\:\mathrm{cos}\:\mathrm{x}\:−\sqrt{\mathrm{sin}\:^{\mathrm{5}} \mathrm{x}}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:\sqrt{\mathrm{sin}\:\mathrm{x}} \\ $$

Answered by john_santu last updated on 14/May/21

(√(sin x)) ≥ 0 ⇒(√(sin x)) cos x(1−sin^4 x−cos^2 x)=0 (1) (√(sin x)) = 0 →x = 2nπ (2) cos x = 0→x=± (π/2)+2nπ (3)1−(1−cos^2 x)^2 −cos^2 x =0 ⇒(1−cos^2 x)(1−(1−cos^2 x))=0 ⇒(1−cos^2 x)cos^2 x = 0 ⇒cos x = 1, →x=2nπ ⇒cos x=−1,→x=π+2nπ

$$\sqrt{\mathrm{sin}\:{x}}\:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\sqrt{\mathrm{sin}\:{x}}\:\mathrm{cos}\:{x}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{4}} \:{x}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\sqrt{\mathrm{sin}\:{x}}\:=\:\mathrm{0}\:\rightarrow{x}\:=\:\mathrm{2}{n}\pi \\ $$$$\left(\mathrm{2}\right)\:\mathrm{cos}\:{x}\:=\:\mathrm{0}\rightarrow{x}=\pm\:\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi \\ $$$$\left(\mathrm{3}\right)\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{cos}\:^{\mathrm{2}} {x}\:=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)\mathrm{cos}\:^{\mathrm{2}} {x}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:{x}\:=\:\mathrm{1},\:\rightarrow{x}=\mathrm{2}{n}\pi \\ $$$$\Rightarrow\mathrm{cos}\:{x}=−\mathrm{1},\rightarrow{x}=\pi+\mathrm{2}{n}\pi\: \\ $$

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