How do you solve the following trigonometric equation?

Question

#cos^2 + 2cosx = 0# when #0 <= x < 2pi#

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Answer 1 · 2018-07-16T05:07:58

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

#cos^2x+2cosx=0#

#cosx(cosx+2)=0#

#cosx=0# or #cosx+2=0#

#cosx=0#
#x=pi/2, (3pi)/2#

#cosx+2=0#
#cosx=-2#
#x=cos^(-1)(-2)#
Therefore, no solution as #cos x# is true only for #-1 <= x <= 1#