Find all solutions to the following

$$\sin \theta + \sqrt{\sin \theta} = 0$$

The only variable occurs in a single trigonometric function, $\sin \theta$. Thus, we attempt to solve for this $\sin \theta$. That is to say, we treat this $\sin \theta$ itself as some unknown variable $x$ and try then to solve for that $x$.

Upon making the substitution $x=\sin \theta$, we can see the underlying structure of things:

$$x + \sqrt{x} = 0$$

Just as we would normally solve the above by isolating the radical and then eliminating it by squaring both sides -- we do the same in our trigonometric equation, as seen below. This in turn leads to an equation that is quadratic in terms of $\sin \theta$, which we deal with in the usual way.

$$\begin{array}{rclc} \sin \theta + \sqrt{\sin \theta} &=& 0\\\\ \sin \theta &=& -\sqrt{\sin \theta} & \overset{\normalsize{\textrm{from here we square both sides to eliminate the radical,}}}{\scriptsize{\textrm{being alert that this may create extraneous solutions...}}}\\\\ \sin^2 \theta &=& \sin \theta & \overset{\normalsize{\textrm{note this is quadratic in terms of } \sin \theta, \textrm{ so we pull}}}{\scriptsize{\textrm{ everything to one side to produce a zero on the other}}}\\\\ \sin^2 \theta - \sin \theta &=& 0 & \scriptsize{\textrm{and then we try to solve by factoring}}\\\\ \sin \theta (\sin \theta - 1) &=& 0 & \overset{\normalsize{\textrm{one of these factors must then be zero,}}}{\scriptsize{\textrm{which gives us two simpler equations to solve}}} \end{array}$$

As seen below, these two simpler equations are both "linear" in terms of $\sin \theta$. In one, the sine value is known -- in the other only a trivial step is needed...

$$\begin{array}{rcl|rcl} \sin \theta &=& 0 \quad & \quad \sin \theta - 1 &=& 0\\\\ & & & \sin \theta &=& 1 &\\\\ \end{array}$$

As noted earlier, the act of squaring may have introduced some extraneous solutions. We should check to see if either of these sine values cause a problem in the original equation:

$$\begin{array}{rcl} \textrm{When } \sin \theta = 0, \quad \sin \theta + \sqrt{\sin \theta} = 0 + \sqrt{0} &=& 0\\ \textrm{When } \sin \theta = 1, \quad \sin \theta + \sqrt{\sin \theta} = 1 + \sqrt{1} &\neq& 0 \quad \textrm{(a problem!)} \end{array}$$

Thus, we reject any values of $\theta$ associated with $\sin \theta = 1$, leaving only solutions for $\theta$ where $\sin \theta = 0$. These of course, correspond to the points on the unit circle whose $y$-coordinates (i.e., "heights") are zero, which can be found quickly: $\theta = 0$ and $\theta = \pi$.

Accounting for co-terminal angles, we have the set of all solutions described by

$$\theta = 0 + 2\pi n \, \, , \, \, \pi + 2\pi n \quad \textrm{where } n \textrm{ is an integer}$$

We could stop there -- although doing so would be silly. There is a much more succinct way to describe these solutions, upon taking advantage of symmetry,

$$\theta = \pi n \quad \textrm{where } n \textrm{ is an integer}$$

As an alternate way to approach this problem, one could ignore trying to eliminate the radical, and instead just factor the left side of the equation immediately: $$\begin{array}{rcl} \sin \theta + \sqrt{\sin \theta} &=& 0\\ \sqrt{\sin \theta} (\sqrt{\sin \theta} + 1) &=& 0\\ \end{array}$$

Solving for where each factor was zero, we find

$$\begin{array}{rcl|rcl} \sqrt{\sin \theta} &=& 0 \quad & \quad \sqrt{\sin \theta} + 1 &=& 0\\ \sin \theta &=& 0 & \sqrt{\sin \theta} &=& -1\\ \end{array}$$

$\sqrt{\sin \theta} = -1$ yields no solutions as no principle square root is negative, while $\sin \theta = 0$ yields the same solutions seen above.