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)

Solution

0 = 1 + 2 cos (3 θ)

Solution

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

Degrees

θ = 4 0^{\circ} + 12 0^{\circ} n, θ = 8 0^{\circ} + 12 0^{\circ} n

Solution steps

0 = 1 + 2 cos (3 θ)

Switch sides

1 + 2 cos (3 θ) = 0

Move

1

to the right side

1 + 2 cos (3 θ) = 0

Subtract

1

from both sides

1 + 2 cos (3 θ) - 1 = 0 - 1

Simplify

2 cos (3 θ) = - 1

2 cos (3 θ) = - 1

Divide both sides by

2

2 cos (3 θ) = - 1

Divide both sides by

2

\frac{2 cos ( 3 θ )}{2} = \frac{- 1}{2}

Simplify

cos (3 θ) = - \frac{1}{2}

cos (3 θ) = - \frac{1}{2}

General solutions for

cos (3 θ) = - \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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}

3 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

3

3 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

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

\frac{\frac{2 π}{3}}{3} + \frac{2 πn}{3}

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{9}

\frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{2 π}{9}

= \frac{2 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}

θ = \frac{2 π}{9} + \frac{2 πn}{3}

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}

3 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

3

3 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

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

\frac{\frac{4 π}{3}}{3} + \frac{2 πn}{3}

\frac{\frac{4 π}{3}}{3} = \frac{4 π}{9}

\frac{\frac{4 π}{3}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{4 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{4 π}{9}

= \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{4 π}{9} + \frac{2 πn}{3}

θ = \frac{4 π}{9} + \frac{2 πn}{3}

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}

Graph

Popular Examples

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tan (4 x) = - 3

sin(pix)=-1/2

sin (π x) = - \frac{1}{2}

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

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

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

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

sin(x)= 9/14

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

Frequently Asked Questions (FAQ)

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