What is the general solution for 0=4|sin(6θ)| ?

The general solution for 0=4|sin(6θ)| is θ=(pin)/3 ,θ= pi/6+(pin)/3

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0=4|sin(6θ)|

0 = 4 ∣ sin (6 θ) ∣

θ = \frac{πn}{3}, θ = \frac{π}{6} + \frac{πn}{3}

Degrees

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

Solution steps

0 = 4 ∣ sin (6 θ) ∣

Switch sides

4 ∣ sin (6 θ) ∣ = 0

Divide both sides by

4

4 ∣ sin (6 θ) ∣ = 0

Divide both sides by

4

\frac{4 ∣ sin ( 6 θ ) ∣}{4} = \frac{0}{4}

Simplify

∣ sin (6 θ) ∣ = 0

∣ sin (6 θ) ∣ = 0

Apply absolute rule: If

∣ u ∣ = 0

then

u = 0

sin (6 θ) = 0

General solutions for

sin (6 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}

6 θ = 0 + 2 πn, 6 θ = π + 2 πn

6 θ = 0 + 2 πn, 6 θ = π + 2 πn

Solve

6 θ = 0 + 2 πn : θ = \frac{πn}{3}

6 θ = 0 + 2 πn

0 + 2 πn = 2 πn

6 θ = 2 πn

Divide both sides by

6

6 θ = 2 πn

Divide both sides by

6

\frac{6 θ}{6} = \frac{2 πn}{6}

Simplify

θ = \frac{πn}{3}

θ = \frac{πn}{3}

Solve

6 θ = π + 2 πn : θ = \frac{π}{6} + \frac{πn}{3}

6 θ = π + 2 πn

Divide both sides by

6

6 θ = π + 2 πn

Divide both sides by

6

\frac{6 θ}{6} = \frac{π}{6} + \frac{2 πn}{6}

Simplify

θ = \frac{π}{6} + \frac{πn}{3}

θ = \frac{π}{6} + \frac{πn}{3}

θ = \frac{πn}{3}, θ = \frac{π}{6} + \frac{πn}{3}

5 sin^{2} (θ) - 11 sin (θ) + 6 = 0

cos (x) = \frac{11}{14}, sin (x)

sin^{2} (2 x) + cos^{2} (3 x) = 1

sin (θ) = 0.9848

cot^{(2)} (x) + 2 cot (x) + 1 = 0

What is the general solution for 0=4|sin(6θ)| ?
The general solution for 0=4|sin(6θ)| is θ=(pin)/3 ,θ= pi/6+(pin)/3