What is the general solution for 2cos^2(x)+7cos(x)+5=0 ?

The general solution for 2cos^2(x)+7cos(x)+5=0 is x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(x)+7cos(x)+5=0

2 cos^{2} (x) + 7 cos (x) + 5 = 0

x = π + 2 πn

Solution steps

2 cos^{2} (x) + 7 cos (x) + 5 = 0

Solve by substitution

cos (x) = - 1, cos (x) = - \frac{5}{2}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - \frac{5}{2} :

No Solution

Combine all the solutions

x = π + 2 πn

2 sin (x) = - 1, 0 \leq x < 2 π

2 cos^{2} (x) = 3 - 3 sin (x)

sec (3 x) - 4 = 0

6 sin^{2} (θ) + 5 sin (θ) = - 2 sin (θ) - 1

t, x = 5 cos (t)

What is the general solution for 2cos^2(x)+7cos(x)+5=0 ?
The general solution for 2cos^2(x)+7cos(x)+5=0 is x=pi+2pin