How do I solve #2 cos^2 x-sin x+1=0# for #0° <=x<=360°#?

1 Answer
Sasha P.
Sep 28, 2015

#x=pi/2#

Explanation:

#cos^2x=1-sin^2x#
#2(1-sin^2x)-sinx+1=0#
#2-2sin^2x-sinx+1=0#
#-2sin^2x-sinx+3=0#
#2sin^2x+sinx-3=0#
#2sin^2x-4sinx+5sinx+2-2-3=0#
#2sin^2x-4sinx+2+5sinx-2-3=0#
#2(sin^2x-2sinx+1)+5sinx-5=0#
#2(sinx-1)^2+5(sinx-1)=0#
#(sinx-1)(2(sinx-1)+5)=0#
#(sinx-1)(2sinx-2+5)=0#
#(sinx-1)(2sinx+3)=0#
#sinx-1=0 vv 2sinx+3=0#
#sinx=1 vv sinx=-3/2#

#x=pi/2# is a solution of the first equation.
#-1<=sinx<=1# so the second equation has no solution.

Finally, #x=pi/2# is a solution.

Related questions
Impact of this question
3157 views around the world
You can reuse this answer
Creative Commons License