Posted by **mike** on Sunday, May 3, 2009 at 9:11pm.

how would you solve for x:

(3)^.5sin(2x)-2cos(2x)+2sin^2(x)=1

using trig identites. please help

- math -
**Reiny**, Sunday, May 3, 2009 at 10:48pm
took me awhile, but here it is ...

√3sin(2x) - 2cos(2x) = 1 - 2sin^2 x

√3sin(2x) - 2cos(2x) = cos(2x)

√3sin(2x) = 3cos(2x)

sin(2x)/cos(2x) = 3/√3

tan(2x) = 3/√3

2x = 60º or 240º

then x = 30º or 120º

or

x = pi/6 or x = 2pi/3

I checked both answers, they work

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