What is the general solution for cos(x)+1/2 =0 ?

The general solution for cos(x)+1/2 =0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(x)+1/2 =0

cos (x) + \frac{1}{2} = 0

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Degrees

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (x) + \frac{1}{2} = 0

Move

\frac{1}{2}

to the right side

cos (x) + \frac{1}{2} = 0

Subtract

\frac{1}{2}

from both sides

cos (x) + \frac{1}{2} - \frac{1}{2} = 0 - \frac{1}{2}

Simplify

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

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

General solutions for

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

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cot (\frac{2 x}{3}) = - 3

sin (x) - 2 sin^{2} (x) = 0

sin (2 θ) + sin (θ) = 0, 0 \leq θ < 2 π

7 sin (x) = cos (x) - 4

5 sin^{2} (θ) - 11 sin (θ) + 2 = 0

What is the general solution for cos(x)+1/2 =0 ?
The general solution for cos(x)+1/2 =0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin