How do you solve $\sin 2\theta = \cos \theta $ ?

Hint: $1.$ Start this question by using the Double angle trigonometric formula for sine.
$2.$ Expand the formula and simplify by adding or subtracting other trigonometric terms.
$3.$ Simplify until it cannot simplify further and until it equals to 0.

 Formula used:
we are going to use Double angle trigonometric formula for sine:
$ \Rightarrow \sin 2\theta = 2\sin \theta \times \cos \theta $

Complete step by step answer:
Firstly, we will be using the double angle formula for sine on the expression given to us:
$ \Rightarrow \sin 2\theta = \cos \theta $
After applying the formula we get:
$ \Rightarrow 2\sin \theta \times \cos \theta = \cos \theta $
Subtract with $\cos \theta $on both the sides of the equation above we get:
$ \Rightarrow 2\sin \theta \times \cos \theta - \cos \theta = \cos \theta - \cos \theta $
Simplify and rewrite the equation:
$ \Rightarrow 2\sin \theta \times \cos \theta - \cos \theta = 0$
Take out the common factor of $\cos \theta $ from left hand side of the equation we get:
$ \Rightarrow \cos \theta \left( {\sin \theta - 1} \right) = 0$
Now, we will evaluate both the terms of left hand side equals to zero and solve them separately:
Equating $\cos \theta $equals to zero we get:
$ \Rightarrow \cos \theta = 0$
From the above expression we can now find the value of $\theta $ the value of cos become zero when:
$ \Rightarrow \theta = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}$……………………. Eq.($2$)
Similarly Equating $2\sin \theta - 1$equals to zero we get:
$ \Rightarrow 2\sin \theta - 1 = 0$
Adding $1$ to both sides of the equation:
$ \Rightarrow 2\sin \theta - 1 + 1 = 0 + 1$
Simplify and rewrite:
$ \Rightarrow 2\sin \theta = 1$
Divided by $2$ on both the sides of the equation:
$ \Rightarrow \dfrac{{2\sin \theta }}{2} = \dfrac{1}{2}$
Simplify and rewrite:
$ \Rightarrow \dfrac{{{2}\sin \theta }}{{{2}}} = \dfrac{1}{2}$
After cancelling we get,
$ \Rightarrow \sin \theta = \dfrac{1}{2}$
From the above expression we can now find the value of $\theta $ the value of sin become 1/2 when:
$ \Rightarrow \theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6}$………………………… eq. $(2)$
From equation 1 and 2 we get four solution $\dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{\pi }{6},\dfrac{{5\pi }}{6}$within the range $0$ to $2\pi $.

Note: Remember that while solving these, you should only change one side of the equation and expand it further.
Before proceeding to a solution, it's important to know the double-angle identity for cosines.
There are three formulas, but since both sides contain sine, we're going to use the formula that includes only sines.
The formula is $\sin 2\theta = 2\sin \theta \times \cos \theta $.