What is the general solution for 5cos(7θ)=0 ?

The general solution for 5cos(7θ)=0 is θ= pi/(14)+(2pin)/7 ,θ=(3pi}{14}+\frac{2pin)/7

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5cos(7θ)=0

Solution

5 cos (7 θ) = 0

Solution

θ = \frac{π}{14} + \frac{2 πn}{7}, θ = \frac{3 π}{14} + \frac{2 πn}{7}

Degrees

θ = 12.85714 \dots^{\circ} + 51.42857 \dots^{\circ} n, θ = 38.57142 \dots^{\circ} + 51.42857 \dots^{\circ} n

Solution steps

5 cos (7 θ) = 0

Divide both sides by

5

5 cos (7 θ) = 0

Divide both sides by

5

\frac{5 cos ( 7 θ )}{5} = \frac{0}{5}

Simplify

cos (7 θ) = 0

cos (7 θ) = 0

General solutions for

cos (7 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}

7 θ = \frac{π}{2} + 2 πn, 7 θ = \frac{3 π}{2} + 2 πn

7 θ = \frac{π}{2} + 2 πn, 7 θ = \frac{3 π}{2} + 2 πn

Solve

7 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{14} + \frac{2 πn}{7}

7 θ = \frac{π}{2} + 2 πn

Divide both sides by

7

7 θ = \frac{π}{2} + 2 πn

Divide both sides by

7

\frac{7 θ}{7} = \frac{\frac{π}{2}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 θ}{7} = \frac{\frac{π}{2}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 θ}{7} : θ

\frac{7 θ}{7}

Divide the numbers:

\frac{7}{7} = 1

= θ

Simplify

\frac{\frac{π}{2}}{7} + \frac{2 πn}{7} : \frac{π}{14} + \frac{2 πn}{7}

\frac{\frac{π}{2}}{7} + \frac{2 πn}{7}

\frac{\frac{π}{2}}{7} = \frac{π}{14}

\frac{\frac{π}{2}}{7}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{π}{14}

= \frac{π}{14} + \frac{2 πn}{7}

θ = \frac{π}{14} + \frac{2 πn}{7}

θ = \frac{π}{14} + \frac{2 πn}{7}

θ = \frac{π}{14} + \frac{2 πn}{7}

Solve

7 θ = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{14} + \frac{2 πn}{7}

7 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

7

7 θ = \frac{3 π}{2} + 2 πn

Divide both sides by

7

\frac{7 θ}{7} = \frac{\frac{3 π}{2}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 θ}{7} = \frac{\frac{3 π}{2}}{7} + \frac{2 πn}{7}

Simplify

\frac{7 θ}{7} : θ

\frac{7 θ}{7}

Divide the numbers:

\frac{7}{7} = 1

= θ

Simplify

\frac{\frac{3 π}{2}}{7} + \frac{2 πn}{7} : \frac{3 π}{14} + \frac{2 πn}{7}

\frac{\frac{3 π}{2}}{7} + \frac{2 πn}{7}

\frac{\frac{3 π}{2}}{7} = \frac{3 π}{14}

\frac{\frac{3 π}{2}}{7}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{3 π}{14}

= \frac{3 π}{14} + \frac{2 πn}{7}

θ = \frac{3 π}{14} + \frac{2 πn}{7}

θ = \frac{3 π}{14} + \frac{2 πn}{7}

θ = \frac{3 π}{14} + \frac{2 πn}{7}

θ = \frac{π}{14} + \frac{2 πn}{7}, θ = \frac{3 π}{14} + \frac{2 πn}{7}

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Frequently Asked Questions (FAQ)

What is the general solution for 5cos(7θ)=0 ?
The general solution for 5cos(7θ)=0 is θ= pi/(14)+(2pin)/7 ,θ=(3pi}{14}+\frac{2pin)/7