What is the general solution for solvefor θ,2cos(θ)+1=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor θ,2cos(θ)+1=0

solve for

θ, 2 cos (θ) + 1 = 0

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

Solution steps

2 cos (θ) + 1 = 0

Move

1

to the right side

2 cos (θ) = - 1

Divide both sides by

2

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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

- 4 + tan (θ) = \frac{- 12 - 3}{3}

2 cos^{2} (x) - 3 = 0

2 sin (x) + 3 tan (x) = 0

tan (t) = 2, 0 < t < \frac{π}{2}, cos (t)

\frac{cos ( x ) + sin ( x )}{sin ( x )} = \frac{1}{tan ( x )}

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