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

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Answer 1 · 2018-04-22T22:40:05

#x = 1.52 , pi,4.78#

#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)