How do you find the general solutions for $\sin x=\cos 2x$?

Hint: Use the formula $\cos 2x=1-2{{\sin }^{2}}x$ to change the cosine function into the terms containing the sine function. Now, form a quadratic equation with the sine function as the variable and use the middle term split method to write the expression as a product of two terms. Substitute each term equal to 0 to find the general solution of the equation. Use the relation: if $\sin x=\sin y$ then $x=n\pi +{{\left( -1 \right)}^{n}}y$ where ‘n’ is any integer.

Complete step by step answer:
Here we have been provided with the trigonometric equation $\sin x=\cos 2x$ and we are asked to find its general solution. To do this, first we need to convert the given equation such that it contains a single trigonometric function. Let us convert them into the equation that contains only the terms of sine function.
$\because \sin x=\cos 2x$
Using the identity $\cos 2x=1-2{{\sin }^{2}}x$ we get,
$\Rightarrow \sin x=1-2{{\sin }^{2}}x$
Taking all the terms to the L.H.S we get,
$\Rightarrow 2{{\sin }^{2}}x+\sin x-1=0$
 Cleary the above relation is a quadratic equation in sine function, so splitting the middle term we get,
$\begin{align}
  & \Rightarrow 2{{\sin }^{2}}x+2\sin x-\sin x-1=0 \\
 & \Rightarrow 2\sin x\left( 1+\sin x \right)-1\left( 1+\sin x \right)=0 \\
 & \Rightarrow \left( 2\sin x-1 \right)\left( 1+\sin x \right)=0 \\
\end{align}$
Substituting each term equal to 0 one by one we get,
(1) Considering $\left( 2\sin x-1 \right)=0$ we get,
$\begin{align}
  & \Rightarrow 2\sin x=1 \\
 & \Rightarrow \sin x=\dfrac{1}{2} \\
\end{align}$
We know that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ so the above equation can be written as:
$\Rightarrow \sin x=\sin \left( \dfrac{\pi }{6} \right)$
Now using the general solution formula for sine function given as: if $\sin x=\sin y$ then $x=n\pi +{{\left( -1 \right)}^{n}}y$ where n is any integer, we get,
$\therefore x=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{6} \right)$, where ‘n’ is any integer.
(2) Considering $\left( 1+\sin x \right)=0$ we get,
$\Rightarrow \sin x=-1$
We know that $\sin \left( \dfrac{\pi }{2} \right)=1$ so we can write the above equation as:
$\Rightarrow \sin x=\sin \left( \dfrac{\pi }{2} \right)$
Again using the formula for the general solution of sine function we get,
$\therefore x=m\pi +{{\left( -1 \right)}^{m}}\left( \dfrac{\pi }{2} \right)$, where ‘m’ is any integer.
Hence the above two cases represent the solution of the given trigonometric equation.

Note: Note that if we have been provided with a particular interval of x in which the solution is to be found then such solutions are called principal solutions. By substituting the integral values of n and m we can find infinite number of solutions of the given equation. The basic thing in the above question is that, to find the solution we have to change different trigonometric functions into a single trigonometric function by applying certain formulas.