Question is solve sin(6x)-cos(3x)=0 from x=0 to 2pi.
Using double angle formula I simplified to 2sin(3x)cos(3x)-cos(3x)=0. I then took out a factor of cos(3x) and got left with cos(3x)=0 or 2sin(3x)-1=0. I solved both of them and overall got a massive 12 solutions which makes me think I've gone wrong somewhere? Like every other question I've ever done has an absolute max of 6. Is this just weird or have I gone wrong?
C4 Trig Equation Watch
- Thread Starter
- 16-02-2016 15:04
- 16-02-2016 16:03
The method is correct. WolframAlpha gives 12 solutions.