math

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how would you solve for x:
(3)^.5sin(2x)-2cos(2x)+2sin^2(x)=1
using trig identites. please help

  • math -

    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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