Question

2cos^2 x - sinx +1 =0

Answer 1

2cos^2 x-sinx +1=0
2(1-sin^2 x) - sin x +1=0
2-2sin ^2 x-sin x +1=0
-2 sin^2 x- sin x +3=0

2sin ^2 x + sin x -3=0
2sin^2 x +3 sinx -2 sin x -3=0
sin x(2 sinx +3)-(2sin x+3)=0
(2 sin x+3)( sin x-1)=0
therefore
2sin x+3=0 or sin x-1=0
sin x= -3/2 or sin x= 1
sin x=1
sinx = sin 90
x=90 degrees

Answer 2

Answer:

Step-by-step explanation:

2cos^2 x-sinx +1=0

2(1-sin^2 x) - sin x +1=0

2-2sin ^2 x-sin x +1=0

-2 sin^2 x- sin x +3=0

2sin ^2 x + sin x -3=0

2sin^2 x +3 sinx -2 sin x -3=0

sin x(2 sinx +3)-(2sin x+3)=0

(2 sin x+3)( sin x-1)=0

therefore

2sin x+3=0 or sin x-1=0

sin x= -3/2 or sin x= 1

sin x=1

sinx = sin 90

x=90 degrees