Compute inverse functions to four significant digits.

cos^2x=3-5cosx

cos^2x + 5cosx -3=0

now you have a quadratic, solve for cos x using the equation

cosx=(-5 -+ sqrt (25 +12)/2

the back of the book says .9987+2kpi and -.09987+2kpi

when i do it i don't get that answer.

I get it, that answer is in radians. The 2kpi is to keep rotation a number of times in the circle.

Thank you so much for helping me I have AC tomorrow and without your help i would have not been able to do that whole section.

You're welcome! I'm glad I could help you understand the solution to the problem. Yes, the answer is in radians because the trigonometric functions such as cosine are typically expressed in radians. And you are correct, the "+2kπ" and "-2kπ" represent the periodic nature of the cosine function, where k is any integer. This indicates that there are infinitely many angles that satisfy the equation.

To obtain the specific inverse functions in radians, you need to evaluate the expression you provided for cos(x).

cos(x) = (-5 ± √(25 + 12)) / 2

cos(x) = (-5 ± √37) / 2

To find the inverse function, you need to find the angles for which cos(x) equals those values.

cos^(-1)((-5 + √37) / 2) ≈ 0.9987 + 2kπ

cos^(-1)((-5 - √37) / 2) ≈ -0.9987 + 2kπ

So, the inverse functions to four significant digits are approximately:

x ≈ 0.9987 + 2kπ

x ≈ -0.9987 + 2kπ

Remember to substitute k with any integer to obtain other valid solutions within the given range. Good luck with your AC!