What is the general solution for -5sin(x)=0,0<= x<= 2pi ?

The general solution for -5sin(x)=0,0<= x<= 2pi is x=0,x=pi,x=2pi

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-5sin(x)=0,0<= x<= 2pi

- 5 sin (x) = 0, 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

Degrees

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

Solution steps

- 5 sin (x) = 0, 0 \leq x \leq 2 π

Divide both sides by

- 5

- 5 sin (x) = 0

Divide both sides by

- 5

\frac{- 5 sin ( x )}{- 5} = \frac{0}{- 5}

Simplify

sin (x) = 0

sin (x) = 0

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

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

3 sec^{2} (θ) - 4 = 0

(cot (x) + 1) (3 tan (x) + 1) = 0

0 = 2 cos (x) - 1

4 sin (2 x) = 2

3 sin^{2} (x) = - 7 sin (x) - 4, 0 \leq x \leq 2 π

What is the general solution for -5sin(x)=0,0<= x<= 2pi ?
The general solution for -5sin(x)=0,0<= x<= 2pi is x=0,x=pi,x=2pi