Solve 2 sin(2 theta + pi/4) - 1 = 0 for theta in the interval from 0 up to but not including 2 pi.

Step-by-step solution

1. 1

   Isolate the sine.

   $\sin\left(2\theta+\frac{\pi}{4}\right)=\frac{1}{2}.$

2. 2

   Substitute for the whole argument and transform the interval. Let $t=2\theta+\tfrac{\pi}{4}$. As $\theta$ runs over $[0,2\pi)$, $2\theta$ runs over $[0,4\pi)$ and so

   $\frac{\pi}{4}\le t<4\pi+\frac{\pi}{4}=\frac{17\pi}{4}.$

   Transforming the interval is the step people skip — the doubled argument sweeps two full turns, so expect roughly twice as many solutions as usual.

3. 3

   Find every t in that stretched interval with sin t = 1/2. In one period the solutions are $t=\tfrac{\pi}{6}$ and $t=\tfrac{5\pi}{6}$; adding multiples of $2\pi$ and keeping those in $\left[\tfrac{\pi}{4},\tfrac{17\pi}{4}\right)$:

   $t=\frac{5\pi}{6},\ \frac{13\pi}{6},\ \frac{17\pi}{6},\ \frac{25\pi}{6}.$

   Note $t=\tfrac{\pi}{6}$ is excluded because $\tfrac{\pi}{6}<\tfrac{\pi}{4}$ — the shift really does drop one solution off the front.

4. 4

   Undo the substitution. From $t=2\theta+\tfrac{\pi}{4}$ we get $\theta=\dfrac{t-\pi/4}{2}$. Taking a common denominator of $12$ inside:

   $\frac{5\pi}{6}-\frac{\pi}{4}=\frac{7\pi}{12},\quad \frac{13\pi}{6}-\frac{\pi}{4}=\frac{23\pi}{12},\quad \frac{17\pi}{6}-\frac{\pi}{4}=\frac{31\pi}{12},\quad \frac{25\pi}{6}-\frac{\pi}{4}=\frac{47\pi}{12}.$

5. 5

   Halve each to get theta.

   $\theta=\frac{7\pi}{24},\ \frac{23\pi}{24},\ \frac{31\pi}{24},\ \frac{47\pi}{24}.$

   All four lie in $[0,2\pi)$, since $\tfrac{47\pi}{24}<\tfrac{48\pi}{24}=2\pi$ ✓.

6. 6

   Check one solution. At $\theta=\tfrac{7\pi}{24}$: $2\theta+\tfrac{\pi}{4}=\tfrac{7\pi}{12}+\tfrac{3\pi}{12}=\tfrac{10\pi}{12}=\tfrac{5\pi}{6}$, and $2\sin\tfrac{5\pi}{6}-1=2(\tfrac12)-1=0$ ✓. The consecutive solutions differ alternately by $\tfrac{16\pi}{24}$ and $\tfrac{8\pi}{24}$, the pattern you expect from two solutions per period repeated twice.