What is the general solution for sin^2(θ)+cos(2θ)-cos(θ)=0 ?

The general solution for sin^2(θ)+cos(2θ)-cos(θ)=0 is θ=2pin,θ= pi/2+2pin,θ=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^2(θ)+cos(2θ)-cos(θ)=0

sin^{2} (θ) + cos (2 θ) - cos (θ) = 0

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

Solution steps

sin^{2} (θ) + cos (2 θ) - cos (θ) = 0

Rewrite using trig identities

- cos (θ) + cos^{2} (θ) = 0

Solve by substitution

cos (θ) = 1, cos (θ) = 0

cos (θ) = 1 : θ = 2 πn

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

Combine all the solutions

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

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

2 sin^{2} (A) + sin (A) = 0

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

- 2 cos (x) + 4 cos (2 x) = 0

2 tan^{2} (x) - 9 tan (x) + 7 = 0

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