3sin2x - 1/3 sinx =03sin2x−13sinx=0
sin2x = 2sinxcosxsin2x=2sinxcosx so implement this into the equation.
3(2sinxcosx) - 1/3 sinx =03(2sinxcosx)−13sinx=0
6sinxcosx - 1/3 sinx =06sinxcosx−13sinx=0
Rearrange equation.
6sinxcosx = 1/3 sinx6sinxcosx=13sinx
Divide by sinxsinx from both sides.
6cosx = 1/36cosx=13
Make cosxcosx the subject of the equation.
cosx = 1/3 *(1/6) -> cosx = 1/18cosx=13⋅(16)→cosx=118
Do inverse cos to find the angle.
x=cos^-1 (1/18) = 1.5152x=cos−1(118)=1.5152 to 5 s.f
Then do CAST diagram or circle to find the repeating angle.
2pi - 1.5152 =4.76782π−1.5152=4.7678 to 5 s.f
x = 1.52 , 4.78x=1.52,4.78
However, if sin(x)=0sin(x)=0, then x =0, x= pix=0,x=π
x=0x=0 can be ignored, so x=pix=π
(Tip: During an exam, check whether the question is in degrees or radians and then change the settings of the calculator before doing the question)