What is the general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) ?

The general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) is θ=(5pi)/6+pin,θ=pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ)

4 tan^{2} (θ) = - 3 tan (θ) + tan^{2} (θ)

θ = \frac{5 π}{6} + πn, θ = πn

Solution steps

4 tan^{2} (θ) = - 3 tan (θ) + tan^{2} (θ)

Solve by substitution

tan (θ) = - \frac{3}{3}, tan (θ) = 0

tan (θ) = - \frac{3}{3} : θ = \frac{5 π}{6} + πn

tan (θ) = 0 : θ = πn

Combine all the solutions

θ = \frac{5 π}{6} + πn, θ = πn

2 sec^{2} (x) - 3 tan (x) = 11

3 cos (2 x) = - 3

4 cos^{2} (θ) + cos (2 θ) - 7 cos (θ) = - 2

cos (θ) = \frac{4 2}{7}

sec^{2} (θ) + 2 sec (θ) = 8

What is the general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) ?
The general solution for 4tan^2(θ)=-sqrt(3)tan(θ)+tan^2(θ) is θ=(5pi)/6+pin,θ=pin