What is the general solution for 2cos^2(x)=-3sin(x)cos(x) ?

The general solution for 2cos^2(x)=-3sin(x)cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=-0.58800…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - 0.58800 \dots + πn

Solution steps

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

Subtract

- 3 sin (x) cos (x)

from both sides

2 cos^{2} (x) + 3 sin (x) cos (x) = 0

Factor

2 cos^{2} (x) + 3 sin (x) cos (x) : cos (x) (2 cos (x) + 3 sin (x))

cos (x) (2 cos (x) + 3 sin (x)) = 0

Solving each part separately

cos (x) = 0 or 2 cos (x) + 3 sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

2 cos (x) + 3 sin (x) = 0 : x = arctan (- \frac{2}{3}) + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arctan (- \frac{2}{3}) + πn

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - 0.58800 \dots + πn

4 cos^{2} (x) = sin^{2} (x) + 3

2 cos^{2} (x) - sin^{2} (x) - 2 sin (x) = 0

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

cos (2 x) + cos (2 x) + 1 = 0

(cos^{2} (x) + cos (x)) \cdot (sin (x) + sin^{3} (x)) = 0

What is the general solution for 2cos^2(x)=-3sin(x)cos(x) ?
The general solution for 2cos^2(x)=-3sin(x)cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=-0.58800…+pin