Question

solve for all values in the interval [0,2pi) 4sin^2x+8cosx-7=0

Answer 1

So, let's first try to put the equation into a quadratic form, and solve that.

Recall that sin^2(x) = 1 - cos^2(x)
So, let's perform that replacement.

4(sin^2(x)) + 8cos(x) - 7 = 4(1-cos^2(x)) + 8cos(x) - 7
= -4cos^2(x) + 8cos(x) - 3.

So, -4cos^2(x) + 8cos(x) - 3 = 0

Let u = cos(x), so then:

-4u^2 + 8u - 3 = 0

Does that look like anything familiar? It's a quadratic equation. This can be solved for u. So, do that, then recall that u = cos(x). I'll let you try to do this, and give help as needed.

Answer 2

use a substitution from

sin^2 + cos^2 = 1

so

sin^2 = 1 - cos^2

put this into your problem and you get

4(1 - cos^2(x)) + 8cos(x) - 7 = 0

and then solve the quadratic