Find the exact solutions in the interval 0 ≤ 𝑥 < 2𝜋 . cos^2𝑥 + 4 = 2 sin 𝑥 − 3

Answer 1

I think the last two lines of Mark's answer should read:

sin(x) = -4 or sin(x) = 2

Therefore, there are NO SOLUTIONS since, for any real number x, -1 ≤ sin(x) ≤ 1.

Answer 2

cos²x - 2sinx + 7 = 0

(1 - sin²x) - 2sinx + 7 = 0

sin²x + 2sinx - 8 = 0

(sinx + 4)(sinx - 2) = 0

sinx = -4 or sinx = 2

So, there are NO SOLUTIONS since, for any real number x, -1 ≤ x ≤ 1.