T tdotgirl New member Joined Oct 4, 2006 Messages 17 Oct 4, 2006 #1 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?
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?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Oct 4, 2006 #2 Hello, tdotgirl! \(\displaystyle 2\sin3x\,-\,1 \:= \:0\) \(\displaystyle 2\sin3x \:= \:1\;\;\Rightarrow\;\;\sin 3x \:= \:\frac{1}{2}\;\;\) . . . Correct! Click to expand... Then: \(\displaystyle \:3x\:=\:\frac{\pi}{6}\,+\,\2\pi n,\:\frac{\5\pi}{6}\,+\,2\pi n\) Therefore: \(\displaystyle \L\:x\;=\;\frac{\pi}{18}\,+\,\frac{2}{3}\pi n,\:\frac{\5\pi}{18}\,+\,\frac{2}{3}\pi n\)
Hello, tdotgirl! \(\displaystyle 2\sin3x\,-\,1 \:= \:0\) \(\displaystyle 2\sin3x \:= \:1\;\;\Rightarrow\;\;\sin 3x \:= \:\frac{1}{2}\;\;\) . . . Correct! Click to expand... Then: \(\displaystyle \:3x\:=\:\frac{\pi}{6}\,+\,\2\pi n,\:\frac{\5\pi}{6}\,+\,2\pi n\) Therefore: \(\displaystyle \L\:x\;=\;\frac{\pi}{18}\,+\,\frac{2}{3}\pi n,\:\frac{\5\pi}{18}\,+\,\frac{2}{3}\pi n\)
