What is the general solution for 2cos(5θ)=0 ?

The general solution for 2cos(5θ)=0 is θ= pi/(10)+(2pin)/5 ,θ=(3pi}{10}+\frac{2pin)/5

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Popular >

2cos(5θ)=0

Solution

2 cos (5 θ) = 0

Solution

θ = \frac{π}{10} + \frac{2 πn}{5}, θ = \frac{3 π}{10} + \frac{2 πn}{5}

Degrees

θ = 1 8^{\circ} + 7 2^{\circ} n, θ = 5 4^{\circ} + 7 2^{\circ} n

Solution steps

2 cos (5 θ) = 0

Divide both sides by

2

2 cos (5 θ) = 0

Divide both sides by

2

\frac{2 cos ( 5 θ )}{2} = \frac{0}{2}

Simplify

cos (5 θ) = 0

cos (5 θ) = 0

General solutions for

cos (5 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}

5 θ = \frac{π}{2} + 2 πn, 5 θ = \frac{3 π}{2} + 2 πn

5 θ = \frac{π}{2} + 2 πn, 5 θ = \frac{3 π}{2} + 2 πn

Solve

5 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{10} + \frac{2 πn}{5}

5 θ = \frac{π}{2} + 2 πn

Divide both sides by

5

5 θ = \frac{π}{2} + 2 πn

Divide both sides by

5

\frac{5 θ}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 θ}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 θ}{5} : θ

\frac{5 θ}{5}

Divide the numbers:

\frac{5}{5} = 1

= θ

Simplify

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5} : \frac{π}{10} + \frac{2 πn}{5}

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

\frac{\frac{π}{2}}{5} = \frac{π}{10}

\frac{\frac{π}{2}}{5}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{π}{10}

= \frac{π}{10} + \frac{2 πn}{5}

θ = \frac{π}{10} + \frac{2 πn}{5}

θ = \frac{π}{10} + \frac{2 πn}{5}

θ = \frac{π}{10} + \frac{2 πn}{5}

Solve

5 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{10} + \frac{2 πn}{5}

5 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

5

5 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

5

\frac{5 θ}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 θ}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 θ}{5} : θ

\frac{5 θ}{5}

Divide the numbers:

\frac{5}{5} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} : \frac{3 π}{10} + \frac{2 πn}{5}

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5}

\frac{\frac{3 π}{2}}{5} = \frac{3 π}{10}

\frac{\frac{3 π}{2}}{5}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{3 π}{10}

= \frac{3 π}{10} + \frac{2 πn}{5}

θ = \frac{3 π}{10} + \frac{2 πn}{5}

θ = \frac{3 π}{10} + \frac{2 πn}{5}

θ = \frac{3 π}{10} + \frac{2 πn}{5}

θ = \frac{π}{10} + \frac{2 πn}{5}, θ = \frac{3 π}{10} + \frac{2 πn}{5}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(5θ)=0 ?
The general solution for 2cos(5θ)=0 is θ= pi/(10)+(2pin)/5 ,θ=(3pi}{10}+\frac{2pin)/5