What is the general solution for 2cos^2(2x)+cos(2x)-1=0 ?

The general solution for 2cos^2(2x)+cos(2x)-1=0 is x= pi/6+pin,x=(5pi)/6+pin,x= pi/2+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

2 cos^{2} (2 x) + cos (2 x) - 1 = 0

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn, x = \frac{π}{2} + πn

Solution steps

2 cos^{2} (2 x) + cos (2 x) - 1 = 0

Solve by substitution

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

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

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

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn, x = \frac{π}{2} + πn

cos (θ) = \frac{0}{8}

3 sin (x) + 2 = 1

sin (y) = - 1

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

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

What is the general solution for 2cos^2(2x)+cos(2x)-1=0 ?
The general solution for 2cos^2(2x)+cos(2x)-1=0 is x= pi/6+pin,x=(5pi)/6+pin,x= pi/2+pin