Solve the trig equation. Apparently the answer is No solution because cos cant be more than 2:

Question

anonymous · Answer

$$\huge \cos^{2}x + \cos(2x) = \frac{5}{4}$$

anonymous · Answer

I can't seem to get No Solution though...

istim · Answer

Anyone going to help? I got a 60% in that class.

anonymous · Answer

Hey @saifoo.khan any ideas?

saifoo.khan · Answer

@Study23 , access is awesome with this!

accessdenied · Answer

There should be a solution to this one.

Try using a trig identity for cos(2x) like 2cos^2 x - 1

anonymous · Answer

@AccessDenied I did. But it didn't work...

accessdenied · Answer

Could you post your process?

anonymous · Answer

Move the 5/4 over to the left and make it cos^2x+cos(2x)-5/4=0 and you can use quadratic equation after that

accessdenied · Answer

Sorry, was hoping we could slip in just before the break!

I used the identity $cos(2x) = cos^2 (x) - 1$.  The rest of the process is mostly Algebra: getting the $cos^2 x$ isolated, taking the positive and negative square root of both sides, and then using the inverse to get $x$.

You can't use it in the form of a quadratic because the linear term is a $cos(2x)$ and the leading term is a $(cos(x))^2$.

anonymous · Answer

@AccessDenied I used the Quadratic formula... so that isnt correct?

accessdenied · Answer

oops i posted the identity incorrectly:
  $cos(2x) = 2cos^2 x - 1$
but yeah, quadratic formula only works if you have $cos(x)$ for the terms

campbell_st · Answer

use cos(2x) = 2cos^(x) - 1

so the problem is now

cos^2(x) + 2cos^2(x) - 1 = 5/4

3cos^2(x) - 1 = 5/4

3cos^2(x) = 9/4

cos^2(x) = 3/4

$$\cos(x) = \pm \sqrt{3}/2$$

there are solutions to this problem in each quadrant

I'll assume you are working in radians

$$\cos^{-1}(\sqrt{3}/2) = pi /3$$