Me plss
cos^2x + sin x + 1 = 0. find x. write down your steps

Step-by-step explanation:

we know : all reals x : (cosx)² + (sinx)² = 1

so : (cosx)² = 1 - (sinx)²

put the value of (cosx)² in equation : (cosx)²+sinx +1 =0

you have :  1 - (sinx)² + sinx +1 = 0

multiply by -1 : (sinx)² - sinx -2 = 0

let : sinx = t     Without forgetting     -1 ≤ t ≤ 1

t² -t -2 = 0  quadrtic equation when  : a = 1        b = -1       c = -2

Δ = b² - 4ac             Δ = (-1)² -4(1)(-2) = 9 = 3²

t1 = (- b +√Δ)/2a   = (1+3)/2 = 2 refused

t2 = (- b -√Δ)/2a   = (1-3)/2 =-1 accept because :  -1 ≤ t ≤ 1

but : sinx = t  so : sinx = - 1

we know  sin a = sinb equi : a = b+2kπ    or a = π - b +2kπ    k in Z(*)

look : -1 = - sin(π/2) = sin(- π/2)   because sin(-c) = - sin(c)

so : sinx = sin(- π/2)

by (*) : x = - π/2 + 2kπ  or x =π -(- π/2) + 2kπ

conclusion : x = - π/2 + 2kπ  or x =3π/2 + 2kπ k in Z

i believe that the answer is

b. y-1=-2(x-4)

hope this !