Question

a) Find all solutions of the equation.
b) Find the solutions in the interval [0, 2pi),

Answer 1

\[2\sin3\theta+1=0\]

Answer 2

a is impossible. in normal terms, because there are infinite solutions.

Answer 3

2 sin (3 theta) +1 = 0
sin (3 theta) = -1/2 sin^-1 both sides
3 theta = -pi/6
\[\theta = -\frac{\pi}{18}\]

All solutions can be given in general terms as:
\[\theta=\frac{\pi}{18}(12n-1)\space\space\space n\in\mathbb{Z}\]
And
\[\theta=\frac{\pi}{18}(12n+7)\space\space\space n\in\mathbb{Z}\]