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

The general solution for 2cos(3θ)=-sqrt(3) is θ=(5pi}{18}+(2pin)/3 ,θ=(7pi)/(18)+\frac{2pin)/3

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2cos(3θ)=-sqrt(3)

2 cos (3 θ) = - 3

θ = \frac{5 π}{18} + \frac{2 πn}{3}, θ = \frac{7 π}{18} + \frac{2 πn}{3}

Solution steps

2 cos (3 θ) = - 3

Divide both sides by

2

cos (3 θ) = - \frac{3}{2}

General solutions for

cos (3 θ) = - \frac{3}{2}

3 θ = \frac{5 π}{6} + 2 πn, 3 θ = \frac{7 π}{6} + 2 πn

Solve

3 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{18} + \frac{2 πn}{3}

Solve

3 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{18} + \frac{2 πn}{3}

θ = \frac{5 π}{18} + \frac{2 πn}{3}, θ = \frac{7 π}{18} + \frac{2 πn}{3}

cos (θ) = - cos (θ)

cot (x) cos (x) = cos (x)

tan^{4} (θ) - 20 tan^{2} (θ) + 64 = 0

sin (2 t) = - cos (2 t)

cos (3 θ) = sin (3 θ)

What is the general solution for 2cos(3θ)=-sqrt(3) ?
The general solution for 2cos(3θ)=-sqrt(3) is θ=(5pi}{18}+(2pin)/3 ,θ=(7pi)/(18)+\frac{2pin)/3