What is the general solution for 8cos(2θ)=-24cos(θ)-16 ?

The general solution for 8cos(2θ)=-24cos(θ)-16 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8cos(2θ)=-24cos(θ)-16

8 cos (2 θ) = - 24 cos (θ) - 16

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

Solution steps

8 cos (2 θ) = - 24 cos (θ) - 16

Subtract

- 24 cos (θ) - 16

from both sides

8 cos (2 θ) + 24 cos (θ) + 16 = 0

Rewrite using trig identities

8 + 16 cos^{2} (θ) + 24 cos (θ) = 0

Solve by substitution

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

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - 1 : θ = π + 2 πn

Combine all the solutions

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

16 cos^{2} (θ) - 9 = 0

sin (a) + cos (a) = 1

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

10 cos (2 x) = 0

6 cos^{2} (θ) sin (θ) - 3 cos^{2} (θ) = 0

What is the general solution for 8cos(2θ)=-24cos(θ)-16 ?
The general solution for 8cos(2θ)=-24cos(θ)-16 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=pi+2pin