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

1 Answer
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.