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

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

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2|sin(4x)|=0

2 ∣ sin (4 x) ∣ = 0

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

Degrees

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

Solution steps

2 ∣ sin (4 x) ∣ = 0

Divide both sides by

2

2 ∣ sin (4 x) ∣ = 0

Divide both sides by

2

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

Simplify

∣ sin (4 x) ∣ = 0

∣ sin (4 x) ∣ = 0

Apply absolute rule: If

∣ u ∣ = 0

then

u = 0

sin (4 x) = 0

General solutions for

sin (4 x) = 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 x = 0 + 2 πn, 4 x = π + 2 πn

4 x = 0 + 2 πn, 4 x = π + 2 πn

Solve

4 x = 0 + 2 πn : x = \frac{πn}{2}

4 x = 0 + 2 πn

0 + 2 πn = 2 πn

4 x = 2 πn

Divide both sides by

4

4 x = 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{2 πn}{4}

Simplify

x = \frac{πn}{2}

x = \frac{πn}{2}

Solve

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

4 x = π + 2 πn

Divide both sides by

4

4 x = π + 2 πn

Divide both sides by

4

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

Simplify

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

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

sin (α) = - \frac{3}{5}

sin (2 x) = \frac{2 \cdot ( 1 \cdot 1 0 ^{12} )}{( 1 \cdot 1 0 ^{7} ) ^{2}}

1 + sin (θ) = \frac{( 3 + 2 )}{2}

- cos (θ) = - cos (2 θ)

cos (x) = \frac{4}{3.3 8 ^{2} + 4 ^{2}}

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