How do you solve for x in the equation 3sin2x-1/3sinx=0 and x in in the interval of [0,2pi)?

1 Answer
Apr 22, 2018

x = 1.52 , pi,4.78x=1.52,π,4.78

Explanation:

3sin2x - 1/3 sinx =03sin2x13sinx=0

sin2x = 2sinxcosxsin2x=2sinxcosx so implement this into the equation.

3(2sinxcosx) - 1/3 sinx =03(2sinxcosx)13sinx=0

6sinxcosx - 1/3 sinx =06sinxcosx13sinx=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=cos1(118)=1.5152 to 5 s.f

Then do CAST diagram or circle to find the repeating angle.

2pi - 1.5152 =4.76782π1.5152=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)