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

The general solution for -2sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-2sin(x)+1=0

- 2 sin (x) + 1 = 0

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

- 2 sin (x) + 1 = 0

Move

1

to the right side

- 2 sin (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

- 2 sin (x) = - 1

- 2 sin (x) = - 1

Divide both sides by

- 2

- 2 sin (x) = - 1

Divide both sides by

- 2

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

Simplify

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

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

General solutions for

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (2 θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

2 - 2 cos^{2} (\frac{x}{2}) = 2 sin^{2} (x)

3 tanh^{2} (x) = 5 sech (x) + 1

tan (θ) = \frac{5}{7}

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

What is the general solution for -2sin(x)+1=0 ?
The general solution for -2sin(x)+1=0 is x= pi/6+2pin,x=(5pi)/6+2pin