SOLUTION: Hi, I was having trouble with this question about trigonometry. Thanks! {{{("cos"^2x+2)/("cos"^2x - 1) + 2(("cos"^2x-1)/("cos"^2x +2)) = 3}}}

Algebra ->  Trigonometry-basics -> SOLUTION: Hi, I was having trouble with this question about trigonometry. Thanks! {{{("cos"^2x+2)/("cos"^2x - 1) + 2(("cos"^2x-1)/("cos"^2x +2)) = 3}}}      Log On


   

Question 1204439: Hi, I was having trouble with this question about trigonometry. Thanks!

Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Let w = cos(x)


turns into
+%28w%5E2+%2B+2%29%2F%28w%5E2+-+1%29+%2B+2%2A%28%28w%5E2+-+1%29%2F%28w%5E2+%2B+2%29%29+=+3

The left hand side simplifies to
%283w%5E4%2B6%29%2F%28w%5E4%2Bw%5E2-2%29+=+3
Let me know if you need to see the steps.

Let's rearrange things a bit
%283w%5E4%2B6%29%2F%28w%5E4%2Bw%5E2-2%29+=+3

3w%5E4%2B6+=+3%28w%5E4%2Bw%5E2-2%29

3w%5E4%2B6+=+3w%5E4%2B3w%5E2-6

6+=+3w%5E2-6 Conveniently the 3w^4 terms cancel

3w%5E2+=+12
It shouldn't be too hard to solve from here.
The two solutions are w = 2 and w = -2

But we run into a problem.
We made w = cos(x)
If w = 2, then cos(x) = 2 which has no real solutions.
Why not? Because the largest cos(x) can get is 1.
The range of cosine is -1+%3C=+y+%3C=+1
Similarly, cos(x) = -2 has no real solutions either.

This will mean the original equation

does not have any real solutions for x.

There is no real number we can substitute in for x, so that the left hand side simplifies to 3.

If you used a tool like WolframAlpha to solve the equation, and clicked "approximate forms", then you'll see the imaginary number i+=+sqrt%28-1%29 is involved.

Another way to verify the "no solutions" is to graph the left hand side (LHS) as one function, and the right hand side (RHS) as another function. The two curves do not intersect.
Here's the desmos graph
https://www.desmos.com/calculator/d0vppysnrr
The LHS curve is entirely below the RHS line.
The highest points on the LHS curve are when y = -3.

Answer by ikleyn(52890) About Me  (Show Source):
You can put this solution on YOUR website!
.

Introduce new variable y = %28cos%5E2%28x%29%2B2%29%2F%28cos%5E2%28x%29+-+1%29+.


Then your equation takes the form

    y + 2%2A%281%2Fy%29 = 3.


Simplify, reducing to quadratic equation

    y^2 - 3y + 2 = 0.


Factor left side

    (y-2)*(y-1) = 0.


The roots are  y= 1  and/or  y= 2.


So now you have two cases.


Case 1.  y= 1  means  %28cos%5E2%28x%29%2B2%29%2F%28cos%5E2%28x%29+-+1%29 = 1,

         and without any further calculations, it is obvious that this equation has no real solutions
        ( because the numerator and the denominator are different, while the right side is 1).



Case 2.  y= 2  means %28cos%5E2%28x%29%2B2%29%2F%28cos%5E2%28x%29+-+1%29 = 2.


         It implies cos^2(x) + 2 = 2*(cos^2(x)-1)

                    cos^2(x) + 2 = 2cos^2(x) - 2

                    2 + 2 = cos^2(x)

                    cos^2(x) = 4,   or  cos(x) = +/- 2,   which OBVIOUSLY has no solutions in real numbers.


ANSWER.  There is NO SOLUTIONS in real numbers.

Solved.