How do you solve \[\sin 2\theta \sin \theta =\cos \theta \]?

Hint: This type of problem is based on the concept of trigonometry. First, we have to consider the given equation. Simplify the left-hand side of the equation using the trigonometric identity \[\sin 2\theta =2\sin \theta \cos \theta \]. Then subtract the whole obtained equation by \[\cos \theta \] so that we get 0 in the right-hand side of the equation. And take \[\cos \theta \] common from the LHS and find the two factors. Now, equate the two factors to zero. We get the value of \[\theta \] using the trigonometric identity, that is \[{{\cos }^{-1}}\left( \cos \theta \right)=\theta \], which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to find the value of \[\theta \] from the equation \[\sin 2\theta \sin \theta =\cos \theta \].
We have been given the equation is \[\sin 2\theta \sin \theta =\cos \theta \]. -----------(1)
We first have to consider the left-hand side of the equation (1).
That is, LHS=\[\sin 2\theta \sin \theta \]
We have to simplify the LHS.
Using the trigonometric identity, that is \[\sin 2\theta =2\sin \theta \cos \theta \], the LHS becomes
LHS=\[\left( 2\sin \theta \cos \theta \right)\sin \theta \]
On further simplifications, we get
LHS=\[2{{\sin }^{2}}\theta \cos \theta \]
Substitute the simplified LHS in the equation (1).
\[\Rightarrow 2{{\sin }^{2}}\theta \cos \theta =\cos \theta \] -------------(2)
Let us now subtract the whole equation (2) by \[\cos \theta \].
Therefore, we get
\[2{{\sin }^{2}}\theta \cos \theta -\cos \theta =\cos \theta -\cos \theta \]
Since terms with same magnitude and opposite signs cancel out, we get
\[2{{\sin }^{2}}\theta \cos \theta -\cos \theta =0\]
Here, we find that \[\cos \theta \] is common in the LHS.
Taking out \[\cos \theta \] common out of the bracket, we get
\[\cos \theta \left( 2{{\sin }^{2}}\theta -1 \right)=0\]
Now, we have found the factors of the given equation.
Since the product of the factors are equal to 0,
Either \[\cos \theta =0\] or \[2{{\sin }^{2}}\theta -1=0\].
Let us consider \[\cos \theta =0\] first.
Take \[{{\cos }^{-1}}\] on both the sides of the equation.
\[\Rightarrow {{\cos }^{-1}}\left( \cos \theta \right)={{\cos }^{-1}}0\].
Using the identity \[{{\cos }^{-1}}\left( \cos \theta \right)=\theta \] in the left-hand side of the equation, we get
\[\theta ={{\cos }^{-1}}0\]
From the trigonometric table, we know that \[\cos \left( \dfrac{n\pi }{2} \right)=0\] where n is the integers.
Therefore, \[\theta =\dfrac{n\pi }{2}\].
Now consider \[2{{\sin }^{2}}\theta -1=0\].
Add 1 on both the sides of the equation.
We get \[2{{\sin }^{2}}\theta -1+1=0+1\].
We know that terms with the same magnitude and opposite signs cancel out.
\[\Rightarrow 2{{\sin }^{2}}\theta =1\]
Divide the above expression by 2.
\[\Rightarrow \dfrac{2{{\sin }^{2}}\theta }{2}=\dfrac{1}{2}\]
Cancelling out the common term 2 from the numerator and denominator of the left-hand side of the equation, we get
\[{{\sin }^{2}}\theta =\dfrac{1}{2}\]
We now have to find the value of \[\sin \theta \].
Take the square root on both sides of the expression.
\[\Rightarrow \sqrt{{{\sin }^{2}}\theta }=\sqrt{\dfrac{1}{2}}\]
We know that \[\sqrt{{{x}^{2}}}=\pm x\]. Therefore, we get
\[\sin \theta =\pm \sqrt{\dfrac{1}{2}}\]
Using the property \[\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\] in the RHS, we get
\[\Rightarrow \sin \theta =\pm \dfrac{\sqrt{1}}{\sqrt{2}}\]
Since \[\sqrt{1}=1\], we get
\[\Rightarrow \sin \theta =\pm \dfrac{1}{\sqrt{2}}\]
From the trigonometric table, we find that
\[\sin \left( \dfrac{n\pi }{4} \right)=\pm \dfrac{1}{\sqrt{2}}\]
Therefore, by comparing the obtained equation with the known formula, we get
\[\theta =\dfrac{n\pi }{4}\], where n is the integer.
Therefore, \[\theta =\dfrac{n\pi }{4},\dfrac{n\pi }{2}\].
Hence, the values of \[\theta \] from the equation \[\sin 2\theta \sin \theta =\cos \theta \] are \[\dfrac{n\pi }{4}\] and \[\dfrac{n\pi }{2}\].

Note: Whenever you get this type of problems, we should simplify the given equation. We should know the trigonometric identities to solve this question. Do not cancel \[\cos \theta \] from both the LHS and RHS directly which will lead to a wrong answer. We should avoid calculation mistakes based on sign conventions.