What is the general solution for 2cos^2(x)+9cos(x)+4=0 ?

The general solution for 2cos^2(x)+9cos(x)+4=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(x)+9cos(x)+4=0

2 cos^{2} (x) + 9 cos (x) + 4 = 0

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solution steps

2 cos^{2} (x) + 9 cos (x) + 4 = 0

Solve by substitution

cos (x) = - \frac{1}{2}, cos (x) = - 4

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - 4 :

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x, arccsc (2 x) = \frac{π}{6}

sin^{2} (x) + tan^{2} (x) = 1

cos (6 x) = sin (x - 1)

tan (θ) = \frac{72}{65}

cos (x) + cos (\frac{x}{2}) + 1 = 0

What is the general solution for 2cos^2(x)+9cos(x)+4=0 ?
The general solution for 2cos^2(x)+9cos(x)+4=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin