What is the general solution for 2sin^2(θ)+3sin(θ)=-1 ?

The general solution for 2sin^2(θ)+3sin(θ)=-1 is θ=(7pi)/6+2pin,θ=(11pi)/6+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 >

2sin^2(θ)+3sin(θ)=-1

2 sin^{2} (θ) + 3 sin (θ) = - 1

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

2 sin^{2} (θ) + 3 sin (θ) = - 1

Solve by substitution

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

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

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

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

tan (4 x) - 3 = 0

cos (3 θ) = - 1, 0 \leq θ < 2 π

2 cos (x) + 3 = 0

cos (θ) = - \frac{4}{9}

cos^{2} (x) + 2 cos (x) = sin^{2} (\frac{x}{2}) - 2

What is the general solution for 2sin^2(θ)+3sin(θ)=-1 ?
The general solution for 2sin^2(θ)+3sin(θ)=-1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin