prelim Q (1 Viewer)

[elliottjohnson]

hey sorry if this is in the wrong thread but its about the subject which isnt a specific thread in the prelim section.

anyway

solve between 0<=x<=360

2sin^2 x-3cosx=2

thanks for any help, if this is in the wrong thread please tell me.

 

[Lukybear]

hey sorry if this is in the wrong thread but its about the subject which isnt a specific thread in the prelim section.

anyway

solve between 0<=x<=360

2sin^2 x-3cosx=2

thanks for any help, if this is in the wrong thread please tell me.

2(1-cos^2x) - 3cosx = 2
2cos^2x + 3cosx = 0
cosx(2cosx+3) = 0
cosx = 0 or -3/2
cosx = 90, 270.

 

[elliottjohnson]

hah, way to go, thats quick

thats the right answer too from the textbook.

how about this one?
0<=x<=360
2sin^2 x + cosx = 1

the answer has four sol'n can u help?

 

[elliottjohnson]

2(1-cos^2x) - 3cosx = 2
2cos^2x + 3cosx = 0
cosx(2cosx+3) = 0
cosx = 0 or -3/2
cosx = 90, 270.

with the sol'n of cosx = -3/2 that has no sol'ns doesnt it cause its over 1??? in the graph?

 

[study-freak]

 

[Drongoski]

2sin²x + cos x = 1

.: 2(1 - cos^2 x) + cos x = 1

.: 2cos^2 x - cos x - 1 = 0

.: (2cos x +1)(cos x -1) = 0

.: cos x = 1 ==> x = 0 or 360 deg

or cos x = - 0.5 ==> x = 120 or 240 deg