How can I solve this?

Question

#2cos^2x +3 cosx =0#

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Answer 1 · 2018-03-14T14:44:17

The solutions are #x=pi/2+2pik,(3pi)/2+2pik#.

#2cos^2x+3cosx=0#

#2(cosx)^2+3cosx=0#

Replace #cosx# with #u#, then solve it like a quadratic:

#2u^2+3u=0#

#u(2u+3)=0#

#u=0,-3/2#

Now, put back in #cosx#:

#cosx=0, color(red)cancelcolor(black)(cosx=-3/2)#

#cosx=0#

Here's a unit circle to remind us where the #x#-values are zero:

enter image source here

#x=pi/2,(3pi)/2#

If you want all of the possible solutions, add #2pik# to each answer, representing that each answer is the same after any multiple of a rotation:

#x=pi/2+2pik,(3pi)/2+2pik#

Answer 2 · 2018-03-14T14:44:26

# x=(2k+1)pi/2, k in ZZ#.

We have, #2cos^2x+3cosx=0#.

#:. cosx(2cosx+3)=0#.

#:. cosx=0, or, cosx=-3/2#.

Since, #AA theta in RR, |costheta|le1, cosx=-3/2" is inadmissible"#.

# cosx=0," gives, "x=(2k+1)pi/2, k in ZZ#.