What is the general solution for 2cos(4θ)=0 ?

The general solution for 2cos(4θ)=0 is θ= pi/8+(pin)/2 ,θ=(3pi)/8+(pin)/2

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2cos(4θ)=0

Solution

2 cos (4 θ) = 0

Solution

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

Degrees

θ = 22. 5^{\circ} + 9 0^{\circ} n, θ = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

2 cos (4 θ) = 0

Divide both sides by

2

2 cos (4 θ) = 0

Divide both sides by

2

\frac{2 cos ( 4 θ )}{2} = \frac{0}{2}

Simplify

cos (4 θ) = 0

cos (4 θ) = 0

General solutions for

cos (4 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}

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

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

Solve

4 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{8} + \frac{πn}{2}

4 θ = \frac{π}{2} + 2 πn

Divide both sides by

4

4 θ = \frac{π}{2} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} : \frac{π}{8} + \frac{πn}{2}

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

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

Solve

4 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{8} + \frac{πn}{2}

4 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

4

4 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} : \frac{3 π}{8} + \frac{πn}{2}

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

\frac{\frac{3 π}{2}}{4} = \frac{3 π}{8}

\frac{\frac{3 π}{2}}{4}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{3 π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{3 π}{8} + \frac{πn}{2}

Popular Examples

1032/3145 sin(4x)+2196/3145 cos(4x)=0

\frac{1032}{3145} sin (4 x) + \frac{2196}{3145} cos (4 x) = 0

6-2sin(θ)-8sin^2(θ)=0

6 - 2 sin (θ) - 8 sin^{2} (θ) = 0

cos(2θ)=cos(3θ)

cos (2 θ) = cos (3 θ)

solvefor θ,2sin(θ)=1.5 solve for

θ, 2 sin (θ) = 1.5

cos(θ)=-2109

cos (θ) = - 2109

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(4θ)=0 ?
The general solution for 2cos(4θ)=0 is θ= pi/8+(pin)/2 ,θ=(3pi)/8+(pin)/2