Question

How do you solve $\sin 2\theta + \cos \theta = 0$ where, $0 < \theta < 2\pi $?

Answer 1

Hint: In this question, we are given a trigonometric equation and the limit between which $\theta $ lies. Start by expanding the first term of the given equation. Take a term common and this will give you two factors. Keep each factor equal to $0$. Then, find the value or values of $\theta $. Use the given limit to find the required values.

Formula used: $\sin 2\theta = 2\sin \theta \cos \theta $

Complete step-by-step solution:
The trigonometric equation given to us is $\sin 2\theta + \cos \theta = 0$.
Firstly, expand the first term of the given equation using a basic trigonometric identity.
$ \Rightarrow \sin 2\theta + \cos \theta = 0$
Using $\sin 2\theta = 2\sin \theta \cos \theta $,
$ \Rightarrow 2\sin \theta .\cos \theta + \cos \theta = 0$
Taking $\cos \theta $ common,
$ \Rightarrow \cos \theta \left( {2\sin \theta + 1} \right) = 0$
Now, we have two factors. We will keep each factor equal to $0$.
$ \Rightarrow \cos \theta = 0,2\sin \theta + 1 = 0$
Next step is to find those angles at which the equation holds true.
First, we will solve the first part. We know that $\cos \theta $ is equal to $0$ at $\dfrac{\pi }{2}$ and $\dfrac{{3\pi }}{2}\left( { = \pi + \dfrac{\pi }{2}} \right)$.
Hence, $\theta = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}$.
Now, we will solve the second part.
$ \Rightarrow 2\sin \theta + 1 = 0$
$ \Rightarrow \sin \theta = \dfrac{{ - 1}}{2}$
We know that $\sin \theta $ is negative in the third and fourth quadrant. Therefore, our required values will also lie in those quadrants only. We know that $\sin \theta $ is equal to $\dfrac{1}{2}$ at $\dfrac{\pi }{6}$ .
Therefore, its negative will be on $\dfrac{{7\pi }}{6}\left( { = \pi + \dfrac{\pi }{6}} \right)$ and $\dfrac{{11\pi }}{6}\left( { = 2\pi - \dfrac{\pi }{6}} \right)$.

Hence, $\theta = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{7\pi }}{6},\dfrac{{11\pi }}{6}$.

Note: What if the range within with $\theta $ lies was not given? Then, till what value would you have found the answer? In such situations, we find a general solution. Each trigonometric ratio has its different general solution. Let us see what would have been our answer, if there had been no range.
$ \Rightarrow \cos \theta = 0,\sin \theta = \dfrac{{ - 1}}{2}$
We can write it as –
$ \Rightarrow \cos \theta = \cos \dfrac{\pi }{2},\sin \theta = \sin \dfrac{{7\pi }}{6}$
Putting them in their general formula –
For $\cos \theta $,
If $\cos \theta = \cos \alpha $, $\theta = \left( {2n + 1} \right)\alpha $.
For $\sin \theta $,
If $\sin \theta = \sin \alpha ,\theta = n\pi + {\left( { - 1} \right)^n}\alpha $.
Now, $\cos \theta = \cos \dfrac{\pi }{2}$.
Therefore, $\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}$.
Now, $\sin \theta = \sin \dfrac{{7\pi }}{6}$.
Therefore, $\theta = n\pi + {\left( { - 1} \right)^n}\dfrac{{7\pi }}{6}$.