What is the general solution for 8cos(θ)-4=0 ?

The general solution for 8cos(θ)-4=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8cos(θ)-4=0

8 cos (θ) - 4 = 0

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solution steps

8 cos (θ) - 4 = 0

Move

4

to the right side

8 cos (θ) = 4

Divide both sides by

8

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

2 (1 + cos (θ)) = 0

0 = cos (π x) \cdot π + sin (π x) π

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

x, tan (x) = - 3

- 9 sin^{2} (x) - 5 cos (x) + 5 = 0

What is the general solution for 8cos(θ)-4=0 ?
The general solution for 8cos(θ)-4=0 is θ= pi/3+2pin,θ=(5pi)/3+2pin