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,4.78

Explanation:

3sin2x - 1/3 sinx =0

sin2x = 2sinxcosx so implement this into the equation.

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

6sinxcosx - 1/3 sinx =0

Rearrange equation.

6sinxcosx = 1/3 sinx

Divide by sinx from both sides.

6cosx = 1/3

Make cosx the subject of the equation.

cosx = 1/3 *(1/6) -> cosx = 1/18

Do inverse cos to find the angle.

x=cos^-1 (1/18) = 1.5152 to 5 s.f

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

2pi - 1.5152 =4.7678 to 5 s.f

x = 1.52 , 4.78

However, if sin(x)=0, then x =0, x= pi

x=0 can be ignored, so x=pi

(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)