# Answer to Question #11809 in Trigonometry for tanya jaiswal

Question #11809

Fin the general solution of

3〖Cos〗^2 θ-√3 SinθCosθ-3〖Sin〗^2 θ=0

3〖Cos〗^2 θ-√3 SinθCosθ-3〖Sin〗^2 θ=0

Expert's answer

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.

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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