Answer to Question #11809 in Trigonometry for tanya jaiswal
Fin the general solution of
3〖Cos〗^2 θ-√3 SinθCosθ-3〖Sin〗^2 θ=0
1
2012-07-17T09:34:32-0400
3(cos^2(x)-sin^2(x))-2sqrt(3)sin(x)cos(x)=0
3cos(2x)-sqrt(3)sin(2x)=0
a) cos2x <> 0
After dividing by cos2x:
tan(2x)=sqrt(3)
2x=π/3
x=π/6
b) If cos2x=0 then 2x=pi/2+pi*k, where any k - integer
It is equivalent to x=pi/4+pi*k/2, where any k - integer.
General solution: x=pi/6 or x=pi/4+pi*k/2, where any k - integer.
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