Solve sin(x) x sin(x) = 1/2 for x, giving the general solution.

Step-by-step solution

1. 1

   Rewrite the product as a square. The left side is $\sin x$ multiplied by itself, so

   $\sin^{2}x=\frac12.$

   Writing it this way matters because it makes the next move — taking a square root, with two signs — obvious.

2. 2

   Take the square root, keeping both signs. From $\sin^{2}x=\tfrac12$,

   $\sin x=\pm\frac{1}{\sqrt2}=\pm\frac{\sqrt2}{2}.$

   Dropping the negative branch would lose half the solutions. The reference angle for $\tfrac{\sqrt2}{2}$ is $\tfrac{\pi}{4}$ (that is $45^{\circ}$), so all solutions sit at $45^{\circ}$ from an axis.

3. 3

   Write the four families, one per quadrant. For $\sin x=+\tfrac{\sqrt2}{2}$ the solutions in $[0,2\pi)$ are $\tfrac{\pi}{4}$ and $\tfrac{3\pi}{4}$; for $\sin x=-\tfrac{\sqrt2}{2}$ they are $\tfrac{5\pi}{4}$ and $\tfrac{7\pi}{4}$. Adding $2k\pi$ to each gives

   $x=\frac{\pi}{4}+2k\pi,\ \frac{3\pi}{4}+2k\pi,\ \frac{5\pi}{4}+2k\pi,\ \frac{7\pi}{4}+2k\pi.$

4. 4

   Merge the four families into one. The four base angles $\tfrac{\pi}{4},\tfrac{3\pi}{4},\tfrac{5\pi}{4},\tfrac{7\pi}{4}$ are equally spaced, each $\tfrac{\pi}{2}$ apart, so the whole solution set collapses to a single arithmetic family:

   $x=\frac{\pi}{4}+\frac{k\pi}{2},\qquad k\in\mathbb{Z}\quad(45^{\circ}+90^{\circ}k).$

   This compact form is both easier to state and much easier to use when the equation is part of a larger problem.

5. 5

   Cross-check with a double-angle identity. Since $\sin^{2}x=\frac{1-\cos 2x}{2}$, the equation is equivalent to

   $\frac{1-\cos 2x}{2}=\frac12\;\Longrightarrow\;\cos 2x=0\;\Longrightarrow\;2x=\frac{\pi}{2}+k\pi\;\Longrightarrow\;x=\frac{\pi}{4}+\frac{k\pi}{2}.$

   The identity route reaches the merged family directly and confirms the answer independently ✓.