Please help...I forget how to do this:

Find all Values of x, in radians, such that 2cos^2x+5cosx -3 =0 where 0 is less than or equel to x which is less than or equel to pi.

Question

Please help...I forget how to do this:

Find all Values of x, in radians, such that 2cos^2x+5cosx -3 =0 where 0 is less than or equel to x which is less than or equel to pi.

anonymous · Answer

Can I use Quadratic equation some how?

raden · Answer

factor out, it can be
2cos^2x+5cosx -3 =0
(2cosx -1)(cosx + 3) = 0
for the zeroes satisfied :
2cosx -1 = 0 or cosx + 3 = 0
continue it ...

anonymous · Answer

2cosx=1
cosx=1/2?

anonymous · Answer

cosx+3 = 0 
cosx = -3

raden · Answer

u knowed that the minimum value for cos x = -1, so there is no solution for cosx=-3. just find the solution of cosx = 1/2

raden · Answer

what angle the cos has value 1/2 ?

anonymous · Answer

60degrees

anonymous · Answer

and 300, but can't be 300 because its 0-pi

anonymous · Answer

Right? Thanks mate! Your a good teacher

raden · Answer

yes.. u r right.
dont forget convert 60 deg to rad :)

anonymous · Answer

1.047 ;) thanks again

raden · Answer

60 degrees = 60/180 pi = pi/3