find all solutions from 0 to 2pi

sin(theta + pi/4) = 1/2

\(sin(\theta + pi/4) = 1/2\)

Looking at just the acute angle.

\(sin\frac{\pi}{6}=\frac{1}{2}\)

sine is positive in the first and second quads so ...

\(\theta+\frac{\pi}{4}=\frac{\pi}{6},\quad \frac{5\pi}{6},\quad \frac{13\pi}{6}\\ \theta+\frac{3\pi}{12}=\frac{2\pi}{12},\quad \frac{10\pi}{12},\quad \frac{26\pi}{12}\\ \theta=\frac{2\pi}{12}-\frac{3\pi}{12},\quad \frac{10\pi}{12}-\frac{3\pi}{12},\quad \frac{26\pi}{12}-\frac{3\pi}{12}\\ \theta=\qquad \frac{-1\pi}{12},\qquad \frac{7\pi}{12},\qquad \frac{23\pi}{12}\\ but\;\;0\le\theta\le2\pi\\ so\\ \theta= \frac{7\pi}{12},\qquad \frac{23\pi}{12}\\ \)

Hi Vest4R

The question is

sin(theta + pi/4) = 1/2

now (theta + pi/4) is an angle so I could substitute    alpha

\(sin\;\alpha=\frac{1}{2}\)

A lot of people just memorize that sin30 degrees=1/2 but I memorize the triangle that it come from. 

I draw an equilateral triangle with sides 2 units and cut it in half.

The right angle triangle thus formed will be this:

Note that 30 degrees is   pi/6 radians

so

\( sin \frac{\pi}{6}=\frac{1}{2}\\ \alpha=\frac{\pi}{6}\\ so\\ \theta+\frac{\pi}{4}=\frac{\pi}{6}\\\)

Does that fully answer your question?    

If you ask more questions you may be advised to private message me this address cause otherwise I may not see it. :)