Solve cos^2(x) = 2 sin(x) -2 on the interval [0,2pi)?

Question

How does one go about solving this problem? I have absolutely no idea how to attack this. Is it an identity?

Answer 1

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

Replace cos^2(x) with 1 - sin^2(x)

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

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

Take sin(x) = u

u^2 + 2u - 3 = 0

Solve for u:

u1 = 1; u2 = -3

Back to sin(x):

sin(x) = u

Since u2 = -3 cannot equal sin(x), we discard that solution

Then

sinx = u1 = 1

sinx = sin(pi/2)

x = pi/2

Answer 2

call your teacher.