Question

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

Answer 1

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.