(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Solve for x. x is greater than or equal to 0 and x is less than or equal to 2pi.

(4/3)sinxcosx-sin^{2}2x=0

If there is a better way to approach this question than the way I did it below, please post it.

Thanks.

2. Relevant equations

N/A

3. The attempt at a solution

(4/3)sinxcosx-sin^{2}2x=0

2(2sinxcosx)-3sin^{2}2x=0

2(sin2x)-3sin^{2}2x=0

2sin2x-3sin^{2}2x=0

sin2x(2-3sinx)=0 <------I know this is incorrect. How do I factor the previous line correctly?

To solve the first part:

sin2x=0

2x=sin^{-1}(0)=0,pi,2pi,3pi,4pi

therefore x=0, pi/2, pi, 3pi/2, 2pi

I know there are other solutions for x. Please help!

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# Homework Help: Urgent: Solve for x in this trigonometric equation?

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