Question

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

Answer 1

Hint: First reduce the given equation in form of $\cos \theta \left\{ 2\sin \theta +1 \right\}=0$. Then use formulas for general equations like $\sin \theta =\dfrac{-1}{2}$ 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

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

For the product of $\cos \theta $ and $\left( 2\sin \theta +1 \right)$ 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 $\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$ where n belongs to integers.

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

We also know that $-\sin \theta =\sin \left( -\theta \right)$ so, $-\sin \dfrac{\pi }{6}=\sin \left( -\dfrac{\pi }{6} \right)$

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

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

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 +{{\left( -1 \right)}^{n}}\theta $ where $\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ ; if $\cos x=\cos \theta $ , then $x=2n\pi \pm \theta $ , where $\theta \in \left[ 0,\pi \right]$ .