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

The general solution for 2sin(x-1)+1=0 is x=(7pi)/6+2pin+1,x=(11pi)/6+2pin+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(x-1)+1=0

2 sin (x - 1) + 1 = 0

x = \frac{7 π}{6} + 2 πn + 1, x = \frac{11 π}{6} + 2 πn + 1

Solution steps

2 sin (x - 1) + 1 = 0

Move

1

to the right side

2 sin (x - 1) = - 1

Divide both sides by

2

sin (x - 1) = - \frac{1}{2}

General solutions for

sin (x - 1) = - \frac{1}{2}

x - 1 = \frac{7 π}{6} + 2 πn, x - 1 = \frac{11 π}{6} + 2 πn

Solve

x - 1 = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{6} + 2 πn + 1

Solve

x - 1 = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{6} + 2 πn + 1

x = \frac{7 π}{6} + 2 πn + 1, x = \frac{11 π}{6} + 2 πn + 1

- 6 sin^{2} (x) + sin (x) = - 3

csc (x) + 2 = 0, 0 \leq x \leq 2 π

tan (x) = 3.7

0 = 12 cos (\frac{π}{3} t) + 8

5 = 20.39 sin (θ) cos (θ)

What is the general solution for 2sin(x-1)+1=0 ?
The general solution for 2sin(x-1)+1=0 is x=(7pi)/6+2pin+1,x=(11pi)/6+2pin+1