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

The general solution for 2cos^2(θ)=2+sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(θ)=2+sin(θ)

2 cos^{2} (θ) = 2 + sin (θ)

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

Solution steps

2 cos^{2} (θ) = 2 + sin (θ)

Subtract

2 + sin (θ)

from both sides

2 cos^{2} (θ) - 2 - sin (θ) = 0

Rewrite using trig identities

- sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = 0

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

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

cos (x) = - cos (x)

3 sin (x) sec (x) = 2 sin (x), 0 \leq x \leq 2 π

2 cos (x) = 3, 0 \leq x \leq 2 π

7 sin (θ) = 5

What is the general solution for 2cos^2(θ)=2+sin(θ) ?
The general solution for 2cos^2(θ)=2+sin(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin,θ=pi+2pin