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

The general solution for 2cos(3θ)=-sqrt(2) is θ= pi/4+(2pin)/3 ,θ=(5pi}{12}+\frac{2pin)/3

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2cos(3θ)=-sqrt(2)

2 cos (3 θ) = - 2

θ = \frac{π}{4} + \frac{2 πn}{3}, θ = \frac{5 π}{12} + \frac{2 πn}{3}

Solution steps

2 cos (3 θ) = - 2

Divide both sides by

2

cos (3 θ) = - \frac{2}{2}

General solutions for

cos (3 θ) = - \frac{2}{2}

3 θ = \frac{3 π}{4} + 2 πn, 3 θ = \frac{5 π}{4} + 2 πn

Solve

3 θ = \frac{3 π}{4} + 2 πn : θ = \frac{π}{4} + \frac{2 πn}{3}

Solve

3 θ = \frac{5 π}{4} + 2 πn : θ = \frac{5 π}{12} + \frac{2 πn}{3}

θ = \frac{π}{4} + \frac{2 πn}{3}, θ = \frac{5 π}{12} + \frac{2 πn}{3}

cos (θ) = 2 sin (θ)

12 sin^{2} (θ) - 3 = 0

2 sin^{2} (θ) + sin (θ) = 0

tan^{2} (x) = 1 - sec^{2} (x)

sec (x) = cos (x)

What is the general solution for 2cos(3θ)=-sqrt(2) ?
The general solution for 2cos(3θ)=-sqrt(2) is θ= pi/4+(2pin)/3 ,θ=(5pi}{12}+\frac{2pin)/3