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Sagot :
Given equation is
[tex]2\sin ^2(x)=1-\cos (x)[/tex][tex]\text{ Use }\sin ^2(x)=1-\cos ^2(x).[/tex][tex]2(1-\cos ^2(x))=1-\cos (x)[/tex][tex]2(1^2-\cos ^2(x))=1-\cos (x)[/tex][tex]\text{ Use }1^2-\cos ^2(x)=(1-\cos (x))(1+\cos (x)).[/tex][tex]2(1-\cos (x))(1+\cos (x))=1-\cos (x)[/tex]Cancel out the common term on both sides, we get
[tex]2(1+\cos (x))=1[/tex]Dividing both sides by 2, we get
[tex]1+\cos (x)=\frac{1}{2}[/tex]Subtracting 1 from both sides, we get
[tex]1+\cos (x)-1=\frac{1}{2}-1[/tex][tex]\cos (x)=-\frac{1}{2}[/tex][tex]\text{Use }\cos (\frac{2\pi}{3})=-\frac{1}{2}\text{.}[/tex][tex]\cos (x)=\cos (\frac{2\pi}{3})[/tex][tex]x=\frac{2\pi}{3}[/tex]Hence the exact value of x is
[tex]x=\frac{2\pi}{3}[/tex]The exact values of all angles in the interval [0, pi] that satisfy the given equation is 2pi/3.
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