Solve 2cos^2x+sinx-1=0, if 0< =x< =2pi

x = π/2, 7π/6, 11π/6

Step-by-step explanation:

2 cos² x + sin x − 1 = 0

Use Pythagorean identity:

2 (1 − sin² x) + sin x − 1 = 0

2 − 2 sin² x + sin x − 1 = 0

-2 sin² x + sin x + 1 = 0

2 sin² x − sin x − 1 = 0

Factor:

(sin x − 1) (2 sin x + 1) = 0

Solve:

sin x = 1 or -1/2

x = π/2, 7π/6, 11π/6

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5

24

step-by-step explanation: