What is the general solution for 3|sin(5θ)|=0 ?

The general solution for 3|sin(5θ)|=0 is θ=(2pin)/5 ,θ= pi/5+(2pin)/5

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3|sin(5θ)|=0

3 ∣ 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

3 ∣ sin (5 θ) ∣ = 0

Divide both sides by

3

3 ∣ sin (5 θ) ∣ = 0

Divide both sides by

3

\frac{3 ∣ sin ( 5 θ ) ∣}{3} = \frac{0}{3}

Simplify

∣ sin (5 θ) ∣ = 0

∣ sin (5 θ) ∣ = 0

Apply absolute rule: If

∣ u ∣ = 0

then

u = 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}

4 cos (x) + 6 = 5 (cos (x) + 1)

cos (x) = 0.8736

4 cos (A) + 3 = 2 cos (A) + 4

sin^{2} (x) - tan (x) = 0

6 sin^{3} (x) + 7 sin^{2} (x) - sin (x) - 2 = 0

What is the general solution for 3|sin(5θ)|=0 ?
The general solution for 3|sin(5θ)|=0 is θ=(2pin)/5 ,θ= pi/5+(2pin)/5