Trig problem...?

Question

2 cos^2= sin x + 1

I'm pretty sure I know what the answer is. So most likely I need the work shown. Because I had this question on my test today and I could only get the answer and of course you need to show your work so I probably got it wrong. So if you could just show the work that would help. Any and all different ways to solve the problem would be most helpful also.

Answer 1

2(1 - sin ² x) = sin x + 1
2 - 2 sin ² x = sin x + 1
2 sin ² x + sinx - 1 = 0
(2 sin x - 1) (sin x + 1) = 0
sin x = 1 / 2 , sin x = - 1
x = 30° , 150° , 270°
or
x = π / 6 , 5π / 6 , 3π / 2

Answer 2

when you're dealing with trigonometric equation always express them in one trigonometric function.
2 cos^2= sin x + 1
2(sin^2 x-1)=sin x+1
2sin^2 x-2 -sin x-1
2sin^2 x - sin x -3=0
(2sin x-3)(sin x+1)=0

2sinx-3=0
x=Sin^-1 (3/2)
no solution since the range of the sine function is from -1 to 1.
sin x+1=0
x=Sin^-1 (-1)= 270 degrees

Answer 3

2 cos^2 x = sin x + 1

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

1 - sin^2 x - sin^2 x = sin x

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

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

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

(Sin x + 1) (2 Sin x - 1) = 0

x = TT or TT/6

Answer 4

2cos²(x) = sin(x) + 1

sin²(x) + cos²(x) = 1
cos²(x) = 1 - sin²(x)

2(1 - sin²(x)) = sin(x) + 1

Let u = sin(x)
2(1-u²) = u + 1
2 - 2u² = u + 1
2u² + u - 1= 0
2u² - u + 2u - 1 = 0
u(2u - 1) + 1(2u - 1) = 0
(u+1)(2u-1) = 0
u = {-1, 1/2}
sin(x) = {-1, 1/2}
x = {(2n+1)π/2, (2n+1)π/2 ± 5π/6} ----- n is an integer

Answer 5

2cos^2 -1 = sin(x)
x = Sin^-1(2(cos^2)-1)