Question

Find the general solution of $\sin 2\theta +\cos \theta =0$ .

Answer 1

Hint: First reduce the given equation in form of $\cos \theta \{2\sin \theta +1\}=0$. Then use formulas for general equations like $\sin \theta ={\displaystyle \frac{-1}{2}}$ and $\cos \theta =0$ to get the desired result.

Complete step-by-step answer:

In the question, we are given an equation which is $\sin 2\theta +\cos \theta =0$ and we have to find general solutions or general values of $\theta$ .

We know that the given equation in question is,

$\sin 2\theta +\cos \theta =0$

Now, we will write $\sin 2\theta$ as $2\sin \theta \cos \theta$ as we know the identity $\sin 2\theta =2\sin \theta \cos \theta$. So, we can write it as

$2\sin \theta \cos \theta +\cos \theta =0$

Now by taking $\cos \theta$ common, we get

$\cos \theta \{2\sin \theta +1\}=0$

For the product of $\cos \theta$ and $(2\sin \theta +1)$ to be ‘0’ either of the two should be 0.

So, we can say that either $\cos \theta =0$ or $2\sin \theta +1=0$ .

Now at first take $\cos \theta =0$ . So, for $\cos \theta$ to be 0 the general value should be $\theta =(2n+1){\displaystyle \frac{\pi }{2}}$ where n belongs to integers.

Now, for the second case $2\sin \theta +1=0$ to be true $\sin \theta$ should be equal to ${\displaystyle \frac{-1}{2}}$. So, $\sin \theta ={\displaystyle \frac{-1}{2}}$ . Now as we know that $\sin {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{1}{2}}$. So, $\sin \theta =-\sin {\displaystyle \frac{\pi }{6}}$

We also know that $-\sin \theta =\sin (-\theta )$ so, $-\sin {\displaystyle \frac{\pi }{6}}=\sin (-{\displaystyle \frac{\pi }{6}})$

Here, $\sin \theta =\sin \left({\displaystyle \frac{-\pi }{6}}\right)$
So, it’s general solution of general value of $\theta$ will be $\theta =n\pi +{(-1)}^n\left({\displaystyle \frac{-\pi }{6}}\right)$ , here n belongs to integers.

Hence, the general solutions are $(2n+1){\displaystyle \frac{\pi }{2}}$ or $x=n\pi +{(-1)}^n\left({\displaystyle \frac{-\pi }{6}}\right)$.

Note: Generally in the questions students confuse like after changing value according to standard angles how to write general solutions so, we use here formula; if $\sin x=\sin \theta$ , then $x=n\pi +{(-1)}^n\theta$ where $\theta \in [-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}]$ ; if $\cos x=\cos \theta$ , then $x=2n\pi \pm \theta$ , where $\theta \in [0,\pi ]$ .