Can anyone help me with a few pre-calculus problems?

Question

1. f(x) = sqrt(9 - x)

How would I find the range for this? Would it just be y is greater than or equal to 0?

2. 2cos^2(x) - cos(x) = 1

I have to solve for x, but I'm not sure where to go other than to factor out a cos(x), leaving the equation:

cos(x) [2cos(x) - 1] = 1

If anyone can help with one or both, that would be great- thanks!

Answer 1

1. Range refers to the possible values of y. Because this is a square root and that can't be negative, your answer is correct.

2. Solve the equation like a quadratic equation. Then use your unit circle to solve for x.

2cos2(x) - cos(x) - 1 = 0

(2cos(x) + 1)(cos(x) - 1) = 0

2cosx + 1 = 0

cosx = -1/2

x = 2pi/3



cosx - 1= 0

cosx = 1

x = 0, 2pi

Answer 2

On 1, you are right.

On 2, when there is something squared, you have to make it equal zero then either reverse FOIL it or do quadratic formula.

2cos^2(x) - cos(x) = 1 subtract 1 from both sides

2 cos² x - cos x - 1 = 0

FOIL: (2 cos x + 1)(cos x - 1) = 0

so either 2 cos x + 1 = 0 making cos x = -1/2

or cos x - 1 = 0 making cos x = 1

so do inverse cosine for those, in the mode (radians or degrees) asked for