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

The general solution for sin(x)+1=0,0<= x<2pi is x=(3pi)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

sin (x) + 1 = 0, 0 \leq x < 2 π

x = \frac{3 π}{2}

Degrees

x = 27 0^{\circ}

Solution steps

sin (x) + 1 = 0, 0 \leq x < 2 π

Move

1

to the right side

sin (x) + 1 = 0

Subtract

1

from both sides

sin (x) + 1 - 1 = 0 - 1

Simplify

sin (x) = - 1

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{2}

θ, sin^{2} (θ) = \frac{1}{4}

cot (θ) = 3, 0 \leq θ < 2 π

(2 cos (2 θ)) = 2

5 sin^{2} (x) = 9 tan (x)

cos (θ) = \frac{5}{4}

What is the general solution for sin(x)+1=0,0<= x<2pi ?
The general solution for sin(x)+1=0,0<= x<2pi is x=(3pi)/2