What is the general solution for 2|sin(4θ)|=0 ?

The general solution for 2|sin(4θ)|=0 is θ=(pin)/2 ,θ= pi/4+(pin)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2|sin(4θ)|=0

2 ∣ sin (4 θ) ∣ = 0

θ = \frac{πn}{2}, θ = \frac{π}{4} + \frac{πn}{2}

Degrees

θ = 0^{\circ} + 9 0^{\circ} n, θ = 4 5^{\circ} + 9 0^{\circ} n

Solution steps

2 ∣ sin (4 θ) ∣ = 0

Divide both sides by

2

2 ∣ sin (4 θ) ∣ = 0

Divide both sides by

2

\frac{2 ∣ sin ( 4 θ ) ∣}{2} = \frac{0}{2}

Simplify

∣ sin (4 θ) ∣ = 0

∣ sin (4 θ) ∣ = 0

Apply absolute rule: If

∣ u ∣ = 0

then

u = 0

sin (4 θ) = 0

General solutions for

sin (4 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}

4 θ = 0 + 2 πn, 4 θ = π + 2 πn

4 θ = 0 + 2 πn, 4 θ = π + 2 πn

Solve

4 θ = 0 + 2 πn : θ = \frac{πn}{2}

4 θ = 0 + 2 πn

0 + 2 πn = 2 πn

4 θ = 2 πn

Divide both sides by

4

4 θ = 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{2 πn}{4}

Simplify

θ = \frac{πn}{2}

θ = \frac{πn}{2}

Solve

4 θ = π + 2 πn : θ = \frac{π}{4} + \frac{πn}{2}

4 θ = π + 2 πn

Divide both sides by

4

4 θ = π + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{π}{4} + \frac{2 πn}{4}

Simplify

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{πn}{2}, θ = \frac{π}{4} + \frac{πn}{2}

250 sin (7 5^{\circ}) = 393.19 sin (4 5^{\circ} - θ)

tan (θ) = \frac{6}{6.71}

x, sin (x) - 1 = 0

sin^{2} (θ) = 1 + cos (2 θ)

4 sin^{2} (θ) = 3 - 2 sin^{2} (θ)

What is the general solution for 2|sin(4θ)|=0 ?
The general solution for 2|sin(4θ)|=0 is θ=(pin)/2 ,θ= pi/4+(pin)/2