Solve this cosine & tangent problem...?

Question

Solve: [ (cosx)^2 - x ][ (tanx)^2 +1 ] = 0 between and including 0 and 2 Pi.

Answer 1

Is this pre-cal or calculus?

Did they give you any picture or formulas to go along with the problem. I remember doing something like this is precal, but its still a little hazy.

I believe you just foil it out like any "normal" problem. But I think cos(x)^2 times tan(x)^2 reduces to something a lot simpler. Look on the inside cover of your book to see the result of multiplying trig functions together.

I think the cos(x) and tan(x) are really cos(theta) and tan(theta). Theta doesn't really matter, they just put it there to represent some angle.  So when I foil everything out I get

cos(theta)^2 times tan(theta)^2 + cos(theta)^2 - xtan(theta)2  -  x = 0

Since I want to get everything with an "x" IN FRONT OF IT, on one side of the equasion, I add  - xtan(theta)2  -  x to the right side of the equation. I am left with this.

cos(theta)^2 times tan(theta)^2 + cos(theta)^2 = xtan(theta)2 + x

I can factor the statement on the right side of the equation out of x, so I do and I am left with this.

cos(theta)^2 times tan(theta)^2 + cos(theta)^2 = x( tan(theta)2 + 1 )

Look on the inside cover of your book to see the result of multiplying trig functions together. It should give you some way to reduce this problem. Sorry I could not totally figure it out, but hope this gives you a good start.