What is the general solution for 0=1+2cos(3θ) ?

The general solution for 0=1+2cos(3θ) is θ=(2pi)/9+(2pin)/3 ,θ=(4pi)/9+(2pin)/3

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0=1+2cos(3θ)

0 = 1 + 2 cos (3 θ)

θ = \frac{2 π}{9} + \frac{2 πn}{3}, θ = \frac{4 π}{9} + \frac{2 πn}{3}

Solution steps

0 = 1 + 2 cos (3 θ)

Switch sides

1 + 2 cos (3 θ) = 0

Move

1

to the right side

2 cos (3 θ) = - 1

Divide both sides by

2

cos (3 θ) = - \frac{1}{2}

General solutions for

cos (3 θ) = - \frac{1}{2}

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

Solve

3 θ = \frac{2 π}{3} + 2 πn : θ = \frac{2 π}{9} + \frac{2 πn}{3}

Solve

3 θ = \frac{4 π}{3} + 2 πn : θ = \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}, θ = \frac{4 π}{9} + \frac{2 πn}{3}

tan (4 x) = - 3

sin (π x) = - \frac{1}{2}

cos^{2} (x) + sin (x) - 1 = 0

- 2 cos^{2} (x) - cos (x) + 1 = 0

sin (x) = \frac{9}{14}

What is the general solution for 0=1+2cos(3θ) ?
The general solution for 0=1+2cos(3θ) is θ=(2pi)/9+(2pin)/3 ,θ=(4pi)/9+(2pin)/3