What is the general solution for 2cos(2θ)+sqrt(3)=0 ?

The general solution for 2cos(2θ)+sqrt(3)=0 is θ=(5pi)/(12)+pin,θ=(7pi)/(12)+pin

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2cos(2θ)+sqrt(3)=0

2 cos (2 θ) + 3 = 0

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

Solution steps

2 cos (2 θ) + 3 = 0

Move

3

to the right side

2 cos (2 θ) = - 3

Divide both sides by

2

cos (2 θ) = - \frac{3}{2}

General solutions for

cos (2 θ) = - \frac{3}{2}

2 θ = \frac{5 π}{6} + 2 πn, 2 θ = \frac{7 π}{6} + 2 πn

Solve

2 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{12} + πn

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

3 sin (x) + sin (2 x) = 0

18 sin^{2} (x) + 27 sin (x) + 9 = 0

3 sin (θ) + 2 sin^{2} (θ) = - 1

cot^{2} (x) + 3 csc (x) = - 3

sec (θ) = 3

What is the general solution for 2cos(2θ)+sqrt(3)=0 ?
The general solution for 2cos(2θ)+sqrt(3)=0 is θ=(5pi)/(12)+pin,θ=(7pi)/(12)+pin