What are the general solutions to the equation?

2cos^2 x  - 5 cos  x  -  3  =  0        factor

(2cos x  + 1)  ( cos x  - 3)   =  0

Set  both factors  to  0   and solve

2cos x + 1  =  0                                   cos x  - 3    =0

2cos x = -1                                          cos x  =  3      [no solution for x here ]

cos x = (-1/2)

This  happens at

x = 2pi /3  + 2pi n         

 and

 x = 4i / 3  +  2 pi n           where n is an integer

What are the general solutions to

\(2cos^2x−5cos\ x−3=0 \)

the equation?

Hello Guest!

\(cos\ x =a \)

\(2a^2-5a-3=0\\ a^2-2.5a-1.5=0\)

\(a=1.25\pm\sqrt{1.25^2+1.5}\\ a=1.25\pm 1.75\\ a_2=3\\ a_1=-0.5\)

\(cos\ x_2=a_2=3\ not\ applicable \)

\(cos\ x_1=a_1=-0.5\\ x=arc\ cos (-0.5) \)

\(\mathbb L_x=\{120^0,240^0\}\)

  !