Trig......?

Question

Solve for X in [0,2pi]

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

I forgot how to do this :(

Answer 1

Let u = cos x

Then

2cos² x + 3cos x - 2 = 0

Becomes

2u² + 3u - 2 = 0

Which factors to

(2u - 1)(u + 2) = 0

So u = 1/2 or u = -2 (but cos x can never = -2, so you can discard that solution).

Re-substitute for u:

cos x = 1/2

And a cosine of 1/2 is associated with 60°, and cos is positive in the 1st and 4th quadrants, so

x = π/3 or 5π/3

Answer 2

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

The first thing to recognize is that this is a quadratic in disguise. Just to show this, I'm going to let u = cos(x). That means we have

2u^2 + 3u - 2 = 0

Now, we factor this as we would a quadratic.

(2u - 1) (u + 2) = 0

This branches off into the equations

2u - 1 = 0 and u + 2 = 0

Which means

u = 1/2
u = -2

Now, we plug u = cos(x) back in, and get

cos(x) = 1/2
cos(x) = -2

We can reject the second equation, since cos(x) will always lie in between -1 and 1. That equation is asking where cos is equal to -2 (i.e. nowhere).

That leaves us with cos(x) = 1/2. This occurs at two values on the unit circle:

x = {pi/3, 5pi/3}

Answer 3

Try factoring... (2cosx -1)(cosx +2), then solve for x