...2cos^2....-5cos.... problem?

Question

2 cos^2 (x) - 5cos(x) -3 = 0

"solve over interval of 0 <= x < 2pi" is what it says

****i dont want the answer****, but can anyone help me get started/going?

its a calc review pack... and dont remember alot of precalc stuff

thanks

Answer 1

Let y = Cos[x]

Then your equation becomes

2 y^2 - 5 y -3 = 0

This is a quadratic equation which yoy can solve for y

There will be two solutions.  In this case only one solution is between

-1 and 1  Call it y1

Then x = ArcCos[y1]