SOLUTION: SecX +tanX= 4 51sinX+34cosx=?

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Question 1103108: SecX +tanX= 4 51sinX+34cosx=?
Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


I had no idea, before starting working on this problem, how to solve it. But by knowing the types of things I CAN do to solve equations like this, I was able to find a way to the solution.

I will try to say something about the reasoning I used at each stage of the solution process as I show the solution itself.

The given equation is
sec%28x%29%2Btan%28x%29+=+4

With my relatively limited experience with trigonometry past the basics, unless the equation is something I recognize, the first thing I do is change everything into sines and cosines:

1%2Fcos%28x%29%2Bsin%28x%29%2Fcos%28x%29+=+4

Next, a general rule in solving equations of almost any type is to clear fractions if possible; so multiply through by cos(x):

1%2Bsin%28x%29+=+4cos%28x%29

Then this is probably the most helpful thing I can tell you about the process of solving this problem, or similar problems with trig equations:
There is little hope in most cases for solving trig equations if the equation involves both sin(x) and cos(x) to the first power.
But there is a good chance it can be solved if the equation contains one or the other, or both, of sin(x) and cos(x), to the second power; generally this will involve the use of the basic trig identity sin%28x%29%5E2%2Bcos%28x%29%5E2+=+1.

With that in mind, let's square both sides of the equation. (And remember that, in doing that, we might introduce extraneous solutions to the original problem.)

1%2B2sin%28x%29%2Bsin%28x%29%5E2+=+16cos%28x%29%5E2%29

Now use the basic trig identity to get the equation in terms of sin(x) only:

1%2B2sin%28x%29%2Bsin%28x%29%5E2+=+16%281-sin%28x%29%5E2%29%29

Now put this quadratic equation in sin(x) in standard form and factor:

1%2B2sin%28x%29%2Bsin%28x%29%5E2+=+16-16sin%28x%29%5E2%29
17sin%28x%29%5E2%2B2sin%28x%29-15=0
%2817sin%28x%29-15%29%28sin%28x%29%2B1%29=0

sin(x) = 15/17 or sin(x) = -1.

Let's check these solutions....

If sin(x) = 15/17, then
cos(x) = 8/17;
sec(x) = 17/8;
tan(x) = 15/8;
sec(x)+tan(x) = 17/8 + 15/8 = 32/8 = 4 IT WORKS!

If sin(x) = -1, then cos(x) is 0, and both sec(x) and tan(x) are undefined. So that solution is extraneous.

And now that we have found the solution to the given equation, it is easy to find the answer to the given question:

51sin%28x%29%2B34cos%28x%29+=+51%2A%2815%2F17%29%2B34%2A%288%2F17%29+=+45%2B16+=+61

Note that, in the solution shown above, I ignored the possibility that, once I found sin(x) = 15/17, cos(x) could be -8/17 instead of 8/17.

In fact I considered that possibility. But with sin(x)=15/17 and cos(x)=-8/17, both sec(x) and tan(x) are negative; and in fact sec(x)+tan(x) = -4, instead of 4. So I rejected that possibility for the value of cos(x).