2 sin3x - 1 = 0

[tdotgirl]

2 sin3x - 1 = 0
2 sin3x = 1
sin 3x = 1/2

sin(2x + x) = 1/2
sin2xcosx + cos2xsinx = 1/2

i'm not sure what to do now.
am i headed in the right direction?

 

[soroban]

Hello, tdotgirl!

\(\displaystyle 2\sin3x\,-\,1 \:= \:0\)

\(\displaystyle 2\sin3x \:= \:1\;\;\Rightarrow\;\;\sin 3x \:= \:\frac{1}{2}\;\;\) . . . Correct!

Then: \(\displaystyle \:3x\:=\:\frac{\pi}{6}\,+\,\2\pi n,\:\frac{\5\pi}{6}\,+\,2\pi n\)

Therefore: \(\displaystyle \L\:x\;=\;\frac{\pi}{18}\,+\,\frac{2}{3}\pi n,\:\frac{\5\pi}{18}\,+\,\frac{2}{3}\pi n\)

 

[tdotgirl]

thank you soroban!!!