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# What are the general solutions to the equation?

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What are the general solutions to the equation?

2cos^2x−5cosx−3=0

Mar 31, 2020

#1
+2

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   Mar 31, 2020

#1
+2

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   CPhill Mar 31, 2020
#2
+2

What are the general solutions to

$$2cos^2x−5cos\ x−3=0$$

the equation?

Hello Guest!

$$cos\ x =a$$

$$2a^2-5a-3=0\\ a^2-2.5a-1.5=0$$

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

$$cos\ x_2=a_2=3\ not\ applicable$$

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

$$\mathbb L_x=\{120^0,240^0\}$$ !

Mar 31, 2020