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

The general solution for 2sin(5θ)=0 is θ=(2pin)/5 ,θ= pi/5+(2pin)/5

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2sin(5θ)=0

2 sin (5 θ) = 0

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

Degrees

θ = 0^{\circ} + 7 2^{\circ} n, θ = 3 6^{\circ} + 7 2^{\circ} n

Solution steps

2 sin (5 θ) = 0

Divide both sides by

2

2 sin (5 θ) = 0

Divide both sides by

2

\frac{2 sin ( 5 θ )}{2} = \frac{0}{2}

Simplify

sin (5 θ) = 0

sin (5 θ) = 0

General solutions for

sin (5 θ) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

5 θ = 0 + 2 πn, 5 θ = π + 2 πn

5 θ = 0 + 2 πn, 5 θ = π + 2 πn

Solve

5 θ = 0 + 2 πn : θ = \frac{2 πn}{5}

5 θ = 0 + 2 πn

0 + 2 πn = 2 πn

5 θ = 2 πn

Divide both sides by

5

5 θ = 2 πn

Divide both sides by

5

\frac{5 θ}{5} = \frac{2 πn}{5}

Simplify

θ = \frac{2 πn}{5}

θ = \frac{2 πn}{5}

Solve

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

5 θ = π + 2 πn

Divide both sides by

5

5 θ = π + 2 πn

Divide both sides by

5

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

Simplify

θ = \frac{π}{5} + \frac{2 πn}{5}

θ = \frac{π}{5} + \frac{2 πn}{5}

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

x, sin^{2} (x) = 1

tan^{4} (x) - 2 = tan^{2} (x) + sec^{2} (x)

cos (x - \frac{π}{4}) = - sin (x)

sin (x) = \frac{5}{13}, cos (x)

tan (θ) = \frac{11.9}{10}

What is the general solution for 2sin(5θ)=0 ?
The general solution for 2sin(5θ)=0 is θ=(2pin)/5 ,θ= pi/5+(2pin)/5