What is the general solution for 2cos(3θ)-1=0 ?

The general solution for 2cos(3θ)-1=0 is θ= pi/9+(2pin)/3 ,θ=(5pi)/9+(2pin)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos(3θ)-1=0

2 cos (3 θ) - 1 = 0

θ = \frac{π}{9} + \frac{2 πn}{3}, θ = \frac{5 π}{9} + \frac{2 πn}{3}

Solution steps

2 cos (3 θ) - 1 = 0

Move

1

to the right side

2 cos (3 θ) = 1

Divide both sides by

2

cos (3 θ) = \frac{1}{2}

General solutions for

cos (3 θ) = \frac{1}{2}

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

Solve

3 θ = \frac{π}{3} + 2 πn : θ = \frac{π}{9} + \frac{2 πn}{3}

Solve

3 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{9} + \frac{2 πn}{3}

θ = \frac{π}{9} + \frac{2 πn}{3}, θ = \frac{5 π}{9} + \frac{2 πn}{3}

cos (4 x) = cos (8 x)

cos (2 x) - cos (x) = 2

4 sin (2 x) + 23 = 0

x, 3 = 18 cos (x)

x, 2 sin (\frac{π ( x - 2 )}{6}) = 1

What is the general solution for 2cos(3θ)-1=0 ?
The general solution for 2cos(3θ)-1=0 is θ= pi/9+(2pin)/3 ,θ=(5pi)/9+(2pin)/3