What is the general solution for 0=4cos(3θ) ?

The general solution for 0=4cos(3θ) is θ= pi/6+(2pin)/3 ,θ= pi/2+(2pin)/3

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0=4cos(3θ)

Solution

0 = 4 cos (3 θ)

Solution

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

Degrees

θ = 3 0^{\circ} + 12 0^{\circ} n, θ = 9 0^{\circ} + 12 0^{\circ} n

Solution steps

0 = 4 cos (3 θ)

Switch sides

4 cos (3 θ) = 0

Divide both sides by

4

4 cos (3 θ) = 0

Divide both sides by

4

\frac{4 cos ( 3 θ )}{4} = \frac{0}{4}

Simplify

cos (3 θ) = 0

cos (3 θ) = 0

General solutions for

cos (3 θ) = 0

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

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

Solve

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

3 θ = \frac{π}{2} + 2 πn

Divide both sides by

3

3 θ = \frac{π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

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

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{2 πn}{3}

θ = \frac{π}{6} + \frac{2 πn}{3}

θ = \frac{π}{6} + \frac{2 πn}{3}

θ = \frac{π}{6} + \frac{2 πn}{3}

Solve

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

3 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

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

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

θ = \frac{π}{2} + \frac{2 πn}{3}

θ = \frac{π}{2} + \frac{2 πn}{3}

θ = \frac{π}{2} + \frac{2 πn}{3}

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

Popular Examples

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sin (θ) = \frac{3}{5}, sin (2 θ)

solvefor x, 1/6 =cos^2(x)solve for

x, \frac{1}{6} = cos^{2} (x)

tan(x)+|tan(x)|=1,-2pi<= ,x<= 2pi

tan (x) + ∣ tan (x) ∣ = 1, - 2 π \leq, x \leq 2 π

2sin^2(x)+3sin(x)=1

2 sin^{2} (x) + 3 sin (x) = 1

11.33=1.59cos(0.99(x-182))+12.14

11.33 = 1.59 cos (0.9 9^{\circ} (x - 18 2^{\circ})) + 12.14

Frequently Asked Questions (FAQ)

What is the general solution for 0=4cos(3θ) ?
The general solution for 0=4cos(3θ) is θ= pi/6+(2pin)/3 ,θ= pi/2+(2pin)/3