Question

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

Answer 1

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 $${\cos }^{-1}(\cos \theta )=\theta$$, 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=$$(2\sin \theta \cos \theta )\sin \theta$$
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
$$\cos \theta (2{\sin }^2\theta -1)=0$$
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.
$$\Rightarrow {\cos }^{-1}(\cos \theta )={\cos }^{-1}0$$.
Using the identity $${\cos }^{-1}(\cos \theta )=\theta$$ in the left-hand side of the equation, we get
$$\theta ={\cos }^{-1}0$$
From the trigonometric table, we know that $$\cos \left({\displaystyle \frac{n\pi }{2}}\right)=0$$ where n is the integers.
Therefore, $$\theta ={\displaystyle \frac{n\pi }{2}}$$.
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.
$$\Rightarrow {\displaystyle \frac{2{\sin }^2\theta }{2}}={\displaystyle \frac{1}{2}}$$
Cancelling out the common term 2 from the numerator and denominator of the left-hand side of the equation, we get
$${\sin }^2\theta ={\displaystyle \frac{1}{2}}$$
We now have to find the value of $$\sin \theta$$.
Take the square root on both sides of the expression.
$$\Rightarrow \sqrt{{\sin }^2\theta }=\sqrt{{\displaystyle \frac{1}{2}}}$$
We know that $$\sqrt{x^2}=\pm x$$. Therefore, we get
$$\sin \theta =\pm \sqrt{{\displaystyle \frac{1}{2}}}$$
Using the property $$\sqrt{{\displaystyle \frac{a}{b}}}={\displaystyle \frac{\sqrt{a}}{\sqrt{b}}}$$ in the RHS, we get
$$\Rightarrow \sin \theta =\pm {\displaystyle \frac{\sqrt{1}}{\sqrt{2}}}$$
Since $$\sqrt{1}=1$$, we get
$$\Rightarrow \sin \theta =\pm {\displaystyle \frac{1}{\sqrt{2}}}$$
From the trigonometric table, we find that
$$\sin \left({\displaystyle \frac{n\pi }{4}}\right)=\pm {\displaystyle \frac{1}{\sqrt{2}}}$$
Therefore, by comparing the obtained equation with the known formula, we get
$$\theta ={\displaystyle \frac{n\pi }{4}}$$, where n is the integer.
Therefore, $$\theta ={\displaystyle \frac{n\pi }{4}},{\displaystyle \frac{n\pi }{2}}$$.
Hence, the values of $$\theta$$ from the equation $$\sin 2\theta \sin \theta =\cos \theta$$ are $${\displaystyle \frac{n\pi }{4}}$$ and $${\displaystyle \frac{n\pi }{2}}$$.

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.