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Question Number 203897 by NasaSara last updated on 01/Feb/24

Answered by deleteduser1 last updated on 01/Feb/24

$${cos}^2\left(x\right)-cos\left(x\right)=\sqrt{1-{cos}^{2}x}$$$$\Rightarrow {cos}^4\left(x\right)+2{cos}^2\left(x\right)-2{cos}^3\left(x\right)-1=0$$$$Forx\in \mathbb{Z}\Rightarrow cos\left(x\right)=1\Rightarrow x=0+2n\pi (n\in \mathbb{Z})$$

Commented by NasaSara last updated on 01/Feb/24

thx

Answered by Frix last updated on 01/Feb/24

$$\mathrm{Obviously}x=2n\pi$$$$$$$$\mathrm{Use}t=\tan \frac{x}{2}\Rightarrow$$$$t(t^3-t^2-t-1)=0$$$$t=0\Rightarrow x=2n\pi$$$$t=\frac{1}{3}+{(\frac{29}{27}+\frac{\sqrt{33}}{9})}^{\frac{1}{3}}+{(\frac{29}{27}-\frac{\sqrt{33}}{9})}^{\frac{1}{3}}$$$$t\approx 1.83929\Rightarrow x\approx 2.14562+2n\pi$$

Answered by Frix last updated on 01/Feb/24

$$\mathrm{Another}\mathrm{weird}\mathrm{path}\mathrm{came}\mathrm{to}\mathrm{mind}...$$$$\cos x=u\wedge \sin x=\sqrt{1-u^2}=\sqrt{1-u}\sqrt{1+u}$$$$\tan \frac{x}{2}=\frac{\sqrt{1-u}}{\sqrt{1+u}}$$$$\frac{{\cos }^{2}x-\cos x-\sin x}{\tan \frac{x}{2}}=\frac{u^2-u-\sqrt{1-u}\sqrt{1+u}}{\frac{\sqrt{1-u}}{\sqrt{1+u}}}=$$$$=-(u+1+u\sqrt{1-u}\sqrt{1+u})$$$$$$$${\cos }^{2}x-\cos x-\sin x=0$$$$-(1+(1+\sin x)\cos x)\tan \frac{x}{2}=0$$$$\tan \frac{x}{2}=0\Rightarrow x=2n\pi$$$$$$$$u+1=-u\sqrt{1-u}\sqrt{1+u}$$$${(u+1)}^2=u^2(1-u)(1+u)$$$$u+1=u^2(1-u)$$$$u^3-u^2+u+1=0$$$$u\approx -.543689=\cos x$$$$x\approx 2.14562+2n\pi$$

Commented by NasaSara last updated on 01/Feb/24

when i put the equation in wolfram alpha some of the solutions where complex ,so how to get the complex solutions ?

Commented by Frix last updated on 02/Feb/24

$$\mathrm{Solve}$$$$t^3-t^2-t-1=0$$$$\mathrm{for}t\in \mathbb{C}$$

Commented by NasaSara last updated on 02/Feb/24

thank you

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