What is the general solution for -9tan^2(θ)+12tan(θ)-4=0 ?

The general solution for -9tan^2(θ)+12tan(θ)-4=0 is θ=0.58800…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-9tan^2(θ)+12tan(θ)-4=0

- 9 tan^{2} (θ) + 12 tan (θ) - 4 = 0

θ = 0.58800 \dots + πn

Solution steps

- 9 tan^{2} (θ) + 12 tan (θ) - 4 = 0

Solve by substitution

tan (θ) = \frac{2}{3}

tan (θ) = \frac{2}{3} : θ = arctan (\frac{2}{3}) + πn

Combine all the solutions

θ = arctan (\frac{2}{3}) + πn

Show solutions in decimal form

θ = 0.58800 \dots + πn

- 5 cos (x) = 0

2 sin^{2} (θ) - sin (θ) = 3

tan (a) = \frac{5}{4}

4 sin (2 x) - 3 = 0

cos (a) = \frac{5}{13}, e = 5 \cdot (tan (a) + sec (a))

What is the general solution for -9tan^2(θ)+12tan(θ)-4=0 ?
The general solution for -9tan^2(θ)+12tan(θ)-4=0 is θ=0.58800…+pin