What is the general solution for cos(2θ)+cos(4θ)=0 ?

The general solution for cos(2θ)+cos(4θ)=0 is θ= pi/6+(2pin)/3 ,θ= pi/2+(2pin)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2θ)+cos(4θ)=0

cos (2 θ) + cos (4 θ) = 0

θ = \frac{π}{6} + \frac{2 πn}{3}, θ = \frac{π}{2} + \frac{2 πn}{3}

Solution steps

cos (2 θ) + cos (4 θ) = 0

Rewrite using trig identities

2 cos (3 θ) cos (θ) = 0

Solving each part separately

cos (3 θ) = 0 or cos (θ) = 0

cos (3 θ) = 0 : θ = \frac{π}{6} + \frac{2 πn}{3}, θ = \frac{π}{2} + \frac{2 πn}{3}

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + \frac{2 πn}{3}, θ = \frac{π}{2} + \frac{2 πn}{3}, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Merge Overlapping Intervals

θ = \frac{π}{6} + \frac{2 πn}{3}, θ = \frac{π}{2} + \frac{2 πn}{3}

6 sin (A) - 3 = 0

2 sin^{2} (2 θ) + 3 sin (2 θ) + 1 = 0

6 cos (2 x) + cos (2 x) = 1

0.8 = \frac{1}{2} (1 - cos (\frac{θ}{2}))

2 tan^{2} (x) + tan (x) - 2 = 0

What is the general solution for cos(2θ)+cos(4θ)=0 ?
The general solution for cos(2θ)+cos(4θ)=0 is θ= pi/6+(2pin)/3 ,θ= pi/2+(2pin)/3