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@Jolly

I will give you some hint.

Code:

sin \left(x \right ) = \cos \left(2x \right ), [0, 2\pi]
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Using \hspace{2 mm} the \hspace{2 mm} identity,\hspace{2 mm} \cos \left(2x \right) = 1 - 2\sin^{2} \left(x \right)
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Hence, \hspace{2 mm} using \hspace{2 mm} the \hspace{2 mm} above \hspace{2 mm} in \hspace{2 mm} the \hspace{2 mm} given \hspace{2 mm} equation.
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sin \left(x \right ) = 1 - 2\sin^{2} \left(x \right)
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\Rightarrow 2\sin^{2} \left(x \right) + sin \left(x \right ) - 1 = 0
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Solving \hspace{2 mm} the \hspace{2 mm} above \hspace{2 mm} quadratic \hspace{2 mm} equation \hspace{2 mm} we \hspace{2 mm} get \hspace{2 mm},
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(2\sin \left(x \right) - 1) (\sin \left(x \right) + 1) = 0

I haven't solved fully but I am giving you a basic idea. If you solve the final equation you will get three solutions for x. Try and see for yourself. Put the above latex code here and convert to see the output.

Sample:

You could also just

let y = sin(x)
2y^2 + y - 1 = 0
So y = 1/2 and y = -1
So sin(x) = 1/2 and sin(x) = -1

And then solve from there using arcsin and a cast diagram/sin graph.