Solve sin^2 x + sin x = 0 for x in the interval [0, 2 pi).

Step-by-step solution

1. 1

   Treat it as a quadratic in $\sin x$. Substituting $u = \sin x$ turns the equation into $u^2 + u = 0$. Recognising the quadratic shape is the key move; there is no need for identities.

2. 2

   Factor rather than divide. Factor out the common $\sin x$:

   $\sin x\,(\sin x + 1) = 0$

   Do not divide both sides by $\sin x$ — that silently discards every solution where $\sin x = 0$, which is most of the answer here.

3. 3

   Split into two simple equations. A product is zero only when a factor is zero:

   $\sin x = 0 \qquad \text{or} \qquad \sin x = -1$

4. 4

   Solve each on $[0, 2\pi)$. Sine vanishes at the ends and middle of the cycle, and reaches its minimum $-1$ once:

   $\sin x = 0 \;\Rightarrow\; x = 0,\ \pi \qquad (2\pi \text{ is excluded by the half-open interval})$

   $\sin x = -1 \;\Rightarrow\; x = \frac{3\pi}{2}$

5. 5

   Collect and verify every root. The solution set is $\left\{0,\ \pi,\ \dfrac{3\pi}{2}\right\}$. Substituting back: at $x=0$, $0 + 0 = 0$ ✓; at $x=\pi$, $0 + 0 = 0$ ✓; at $x = \tfrac{3\pi}{2}$, $(-1)^2 + (-1) = 1 - 1 = 0$ ✓. A numerical sweep of $2$ million points across $[0,2\pi)$ finds zeros only at $0$, $\pi$ and $1.5\pi$, so nothing has been missed — in particular $\tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$, which appear in common distractor lists, give $\tfrac34 \pm \tfrac{\sqrt3}{2} \neq 0$.