What is the general solution for 2cos^2(2x)+3cos(2x)-2=0 ?

The general solution for 2cos^2(2x)+3cos(2x)-2=0 is x= pi/6+pin,x=(5pi)/6+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(2x)+3cos(2x)-2=0

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

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Solution steps

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

Solve by substitution

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

cos (2 x) = \frac{1}{2} : x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

cos (2 x) = - 2 :

No Solution

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

tan (t) = - \frac{12}{5}, \frac{3 π}{2} < t < 2 π

3 sin (x) + 5 cos (x) = 4

- 2 = sec (x)

2 sin^{2} (x) + 4 sin (x) = 2 cos^{2} (x) + 1

y, sec^{2} (y) = x

What is the general solution for 2cos^2(2x)+3cos(2x)-2=0 ?
The general solution for 2cos^2(2x)+3cos(2x)-2=0 is x= pi/6+pin,x=(5pi)/6+pin