Solve, sin⁡(2x)=cos⁡(x), on the interval [0,2π)

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stamptsk · Accepted Answer

Observe that sin⁡(2x)=2sin⁡xcos⁡x, so that  sin⁡(2x)=cos⁡x⇔cos⁡x(2sin⁡x−1)=0⇔cos⁡x=0  or  2sin⁡x−1=0  The final pair of equations is solved in a standard way. The equation cos⁡x=0 has two solutions in [0,2π) namely x=π2,,3π2. Furthermore,  2sin⁡x−1=0⇔sin⁡x=12  which also has two solutions in [0,2π) namely x=π6,,5π6  Hence the solutions to the original equation are x∈{π2,3π2,π6,5π6}

Hana Larsen · Accepted Answer

Observe that sin⁡(2x)=2sin⁡xcos⁡x, so that    sin⁡(2x)=cos⁡x⇔cos⁡x(2sin⁡x−1)=0⇔cos⁡x=0  or  2sin⁡x−1=0    The final pair of equations is solved in a standard way. The equation cos⁡x=0 has two solutions in [0,2π) namely x=π2,,3π2. Furthermore,    2sin⁡x−1=0⇔sin⁡x=12    which also has two solutions in [0,2π) namely x=π6,,5π6    Hence the solutions to the original equation are x∈{π2,3π2,π6,5π6}

Solve, \sin(2x)=\cos(x), on the interval [0,2\pi)

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