What is the general solution for sin(x)=1,0<= x<= 2pi ?

The general solution for sin(x)=1,0<= x<= 2pi is x= pi/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(x)=1,0<= x<= 2pi

sin (x) = 1, 0 \leq x \leq 2 π

x = \frac{π}{2}

Degrees

x = 9 0^{\circ}

Solution steps

sin (x) = 1, 0 \leq x \leq 2 π

General solutions for

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}

6 csc (θ) - 6 = 0

9 sin (x) tan (x) + 4 tan (x) = 0

x, f [g] = cos (\frac{1}{x ^{2}})

2 cos (\frac{π}{5} θ) = 3

2 cos (x) - sin (2 x) = 0

What is the general solution for sin(x)=1,0<= x<= 2pi ?
The general solution for sin(x)=1,0<= x<= 2pi is x= pi/2