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

The general solution for 2sin^2(θ)+cos(θ)-1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(θ)+cos(θ)-1=0

2 sin^{2} (θ) + cos (θ) - 1 = 0

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

Solution steps

2 sin^{2} (θ) + cos (θ) - 1 = 0

Rewrite using trig identities

1 + cos (θ) - 2 cos^{2} (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 = 2 πn

Combine all the solutions

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

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

6 sin^{2} (θ) - sin (θ) - 2 = 0, 0 \leq θ \leq 2 π

33 cot (x) = 3

2 tan^{2} (x) - 6 = 0

0 = 2 sin (x) - 1

What is the general solution for 2sin^2(θ)+cos(θ)-1=0 ?
The general solution for 2sin^2(θ)+cos(θ)-1=0 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin,θ=2pin