What is the general solution for sin(x)=0,-pi/4 <x<(9pi)/4 ?

The general solution for sin(x)=0,-pi/4 <x<(9pi)/4 is x=0,x=pi,x=2pi

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sin(x)=0,-pi/4 <x<(9pi)/4

sin (x) = 0, - \frac{π}{4} < x < \frac{9 π}{4}

x = 0, x = π, x = 2 π

Degrees

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}

Solution steps

sin (x) = 0, - \frac{π}{4} < x < \frac{9 π}{4}

General solutions for

sin (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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

- \frac{π}{4} < x < \frac{9 π}{4}

x = 0, x = π, x = 2 π

2 tan (x) cos (x) = tan (x)

sin (x) = \frac{5}{4}

sin (2 θ) = 3 cos (θ)

x, tan (x) + ∣ tan (x) ∣ = 1

0 = 2 sin^{2} (θ) + 3 sin (θ)

What is the general solution for sin(x)=0,-pi/4 <x<(9pi)/4 ?
The general solution for sin(x)=0,-pi/4 <x<(9pi)/4 is x=0,x=pi,x=2pi