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December 20, 2014

December 20, 2014

Posted by **ashley --- HELP PLEASEE!!!** on Sunday, April 10, 2011 at 10:46am.

- trig -
**drwls**, Sunday, April 10, 2011 at 11:22amUse the equation

sin(2x) = 2 sin x cos x

so that

6 sinx cosx -2 cosx = 0

cosx*(1 - 3sinx) = 0

(I divided out -2)

So the equation is satisfied wherever cosx = 0 or sinx = 1/3

x = pi/2 and 0.3398 radians are two solutions. There are others in other quadrants.

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