How do you solve $\sin 2\theta + \cos \theta = 0$?

Hint:Expand $\sin 2\theta $to simplify the equation and then take out common terms to solve the equationSince the question comprises of two unlike terms we will first try to either convert them into like terms or at least convert it to a more simpler equation. To do this, we will first simplify $\sin 2\theta $with $2\sin \theta \cos $. After that, we will take $\cos \theta $common from both the terms which will leave us either with $\cos \theta = 0$or $2\sin \theta + 1 = 0$. Solving these two will give us two different solutions.

Complete step by step solution:
Here, the given equation is $\sin 2\theta + \cos \theta = 0$ (1)
But we know that,
\[\sin 2\theta = 2\sin \theta \cos \theta \] (2)
Putting value of $\sin 2\theta $from equation 2 in 1 we get,
\[2\sin \theta \cos \theta + \cos \theta = 0\]
Taking \[\cos \theta \]common from the equation
\[\cos \theta (2\sin \theta + 1) = 0\]
Now, we know that whenever the product of any two variables is zero. Either or both of the variable will be equal to zero
\[\therefore \,\,\]either \[\cos \theta = 0\]or \[2\sin \theta + 1 = 0\]
solving these two separately :-
\[
\cos \theta = 0 \\
\Rightarrow \theta = {90^ \circ } \\
\]
and,
$
2\sin \theta + 1 = 0 \\
\Rightarrow 2\sin \theta = - 1 \\
\Rightarrow \sin \theta = \dfrac{{ - 1}}{2} \\
\Rightarrow \theta = \dfrac{{7\pi }}{6} \\
$
Hence, for the given equation $\sin 2\theta + \cos \theta = 0$, we have 2 solution, $\theta = {90^ \circ }$or $\theta = \dfrac{{7\pi }}{6}$
Alternate solution:
You can also find the general solution of the two equation to make the answer more general
Here,
\[\cos \theta = 0\]
Now general solution for \[\cos \theta = 0\]is given by the formula $\theta = \dfrac{{(2n + 1)\pi }}{2}$
and for ,
$
2\sin \theta + 1 = 0 \\
\Rightarrow 2\sin \theta = - 1 \\
\Rightarrow \sin \theta = \dfrac{{ - 1}}{2} \\
\Rightarrow \theta = \dfrac{{7\pi }}{6} \\
$
Now general solution of \[\sin \theta = - k\] is given by \[\theta = n\pi + {( - 1)^n}k\]
therefore, our general solution becomes $\sin \theta = \dfrac{{ - 1}}{2}$$\theta = n\pi + {( -
1)^n}(\dfrac{{7\pi }}{6})$
Hence, two general solution of $\sin 2\theta + \cos \theta = 0$are $\theta = \dfrac{{(2n + 1)\pi }}{2}$ and $\theta = n\pi + {( - 1)^n}(\dfrac{{7\pi }}{6})$.

Note: The solution of $\sin \theta = \dfrac{{ - 1}}{2}$should be calculated carefully. The sign of sin is negative which means you should consider sin in either the 3 rd or 4 th quadrant. Also, the value of \[\cos \theta \]should be calculated carefully as the sign is neither positive or negative but straight equal to 0.