# Thread: Solving a trig multiple angle?

1. ## Solving a trig multiple angle?

I have this expression:

sin^2(3x)=1/4 (reads sine squared three x equals one fourth)

when i solved it i got sinx= (plus or minus) 1/2 and the x values were from (pi)/18 to 35(pi)/18 with a total of 12 possible x values on the interval (0, 6(pi)) . Could anyone verify for me if this is correct? thanks a lot!

2. Hello, nascar77!

I think I got the same answers . . .

$\sin^2(3x)\:=\:\frac{1}{4}$

When i solved it, i got: . $\sin 3x\:=\:\pm\tfrac{1}{2}$

and the $x$ values were from $\frac{\pi}{18}$ to $\frac{35\pi}{18}$

with a total of 12 possible $x$-values on the interval $(0,\:6\pi)$ . . . . Did you mean ${\color{blue}(0,\:2\pi)}$ ?

$\text{We have: }\;\sin(3x) \:=\:\pm\frac{1}{2} \quad\Rightarrow\quad 3x \;=\;\begin{Bmatrix}\dfrac{\pi}{6} + 2\pi n \\ \\[-3mm] \dfrac{5\pi}{6} + 2\pi n \\ \\[-3mm] \dfrac{7\pi}{6} + 2\pi n \\ \\[-3mm] \dfrac{11\pi}{6} + 2\pi n \end{Bmatrix}$

$\text{Therefore: }\;x \;=\;\begin{Bmatrix}\dfrac{\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{5\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{7\pi}{18} + \dfrac{2\pi}{3}n \\ \\[-3mm] \dfrac{11\pi}{18} + \dfrac{2\pi}{3}n \end{Bmatrix}$ . $= \;\begin{pmatrix}\dfrac{\pi}{18} & \dfrac{13\pi}{18} & \dfrac{25\pi}{18} \\ \\[-3mm] \dfrac{5\pi}{18} & \dfrac{17\pi}{18} & \dfrac{29\pi}{18} \\ \\[-3mm] \dfrac{7\pi}{18} & \dfrac{19\pi}{18} & \dfrac{31\pi}{18} \\ \\[-3mm] \dfrac{11\pi}{18} & \dfrac{23\pi}{18} & \dfrac{35\pi}{18} \end{pmatrix}$