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Trigonometry 14 Online
OpenStudy (anonymous):

Solve 2sin^2(2x)=1

OpenStudy (anonymous):

i don't think that is the right approach try \[\sin^2(2x)=\frac{1}{2}\\ \sin(2x)=\pm\frac{\sqrt{2}}{2}\] then solve for \(2x\) and divide by 2 to solve for \(x\)

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