What is the general solution for 4+2tan^2(θ)=5tan^2(θ)+3 ?

The general solution for 4+2tan^2(θ)=5tan^2(θ)+3 is θ=0.52359…+pin,θ=-0.52359…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4+2tan^2(θ)=5tan^2(θ)+3

4 + 2 tan^{2} (θ) = 5 tan^{2} (θ) + 3

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

Solution steps

4 + 2 tan^{2} (θ) = 5 tan^{2} (θ) + 3

Solve by substitution

tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = - \frac{1}{3} : θ = arctan (- \frac{1}{3}) + πn

Combine all the solutions

θ = arctan (\frac{1}{3}) + πn, θ = arctan (- \frac{1}{3}) + πn

Show solutions in decimal form

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

sin (x) = sin (\frac{π}{5})

x, sin (2 x) = cos (2 x)

cos (x) = \frac{16}{15}

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

0 = 2 (1 + cos (θ))

What is the general solution for 4+2tan^2(θ)=5tan^2(θ)+3 ?
The general solution for 4+2tan^2(θ)=5tan^2(θ)+3 is θ=0.52359…+pin,θ=-0.52359…+pin