What is the general solution for 2sec(x)tan(x)+2sec(x)+tan(x)+1=0 ?

The general solution for 2sec(x)tan(x)+2sec(x)+tan(x)+1=0 is x=(3pi)/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sec(x)tan(x)+2sec(x)+tan(x)+1=0

2 sec (x) tan (x) + 2 sec (x) + tan (x) + 1 = 0

x = \frac{3 π}{4} + πn

Solution steps

2 sec (x) tan (x) + 2 sec (x) + tan (x) + 1 = 0

Factor

2 sec (x) tan (x) + 2 sec (x) + tan (x) + 1 : (tan (x) + 1) (2 sec (x) + 1)

(tan (x) + 1) (2 sec (x) + 1) = 0

Solving each part separately

tan (x) + 1 = 0 or 2 sec (x) + 1 = 0

tan (x) + 1 = 0 : x = \frac{3 π}{4} + πn

2 sec (x) + 1 = 0 :

No Solution

Combine all the solutions

x = \frac{3 π}{4} + πn

6 tan^{2} (θ) cos (θ) + 3 tan^{2} (θ) = 0

2 sin^{2} (x) - cos (x) = 1, 0 \leq x < 2 π

cos^{2} (x) + sin (x) = sin^{2} (x)

1 = - cot (x)

1 = sec (x)

What is the general solution for 2sec(x)tan(x)+2sec(x)+tan(x)+1=0 ?
The general solution for 2sec(x)tan(x)+2sec(x)+tan(x)+1=0 is x=(3pi)/4+pin