What is the general solution for 6cos(2θ)+3=0 ?

The general solution for 6cos(2θ)+3=0 is θ= pi/3+pin,θ=(2pi)/3+pin

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6cos(2θ)+3=0

Solution

6 cos (2 θ) + 3 = 0

Solution

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

Degrees

θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

6 cos (2 θ) + 3 = 0

Move

3

to the right side

6 cos (2 θ) + 3 = 0

Subtract

3

from both sides

6 cos (2 θ) + 3 - 3 = 0 - 3

Simplify

6 cos (2 θ) = - 3

6 cos (2 θ) = - 3

Divide both sides by

6

6 cos (2 θ) = - 3

Divide both sides by

6

\frac{6 cos ( 2 θ )}{6} = \frac{- 3}{6}

Simplify

\frac{6 cos ( 2 θ )}{6} = \frac{- 3}{6}

Simplify

\frac{6 cos ( 2 θ )}{6} : cos (2 θ)

\frac{6 cos ( 2 θ )}{6}

Divide the numbers:

\frac{6}{6} = 1

= cos (2 θ)

Simplify

\frac{- 3}{6} : - \frac{1}{2}

\frac{- 3}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{6}

Cancel the common factor:

3

= - \frac{1}{2}

cos (2 θ) = - \frac{1}{2}

cos (2 θ) = - \frac{1}{2}

cos (2 θ) = - \frac{1}{2}

General solutions for

cos (2 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}

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

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

Solve

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

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

Solve

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

2 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

2

2 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

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

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{4 π}{6}

Cancel the common factor:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

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

Popular Examples

(sin(θ)+cos^2(θ))^2=2sin(θ)

(sin (θ) + cos^{2} (θ))^{2} = 2 sin (θ)

2/3 =sin(2θ)

\frac{2}{3} = sin (2 θ)

sin(3x)-2=-3sin(3x)

sin (3 x) - 2 = - 3 sin (3 x)

sin(a)= 1/5

sin (a) = \frac{1}{5}

solvefor θ,cos(θ)= 2/3 solve for

θ, cos (θ) = \frac{2}{3}

Frequently Asked Questions (FAQ)

What is the general solution for 6cos(2θ)+3=0 ?
The general solution for 6cos(2θ)+3=0 is θ= pi/3+pin,θ=(2pi)/3+pin