What is the general solution for -sin(4x)*4=0 ?

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

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-sin(4x)*4=0

- sin (4 x) \cdot 4 = 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

- sin (4 x) \cdot 4 = 0

Divide both sides by

- 4

- sin (4 x) \cdot 4 = 0

Divide both sides by

- 4

\frac{- sin ( 4 x ) \cdot 4}{- 4} = \frac{0}{- 4}

Simplify

sin (4 x) = 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}

cos (θ) = \frac{5}{6}

cos (2 x) = cos^{2} (x) - \frac{1}{2}

cos (x) = 1, 0 \leq x \leq 2 π

x, cot (- x) cos (- x) + sin (- x) = \frac{- 1}{f}

tan^{2} (x) sin (x) = 3 sin (x)

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