<PRE><B>Question 1103108</B>
&lt;br&gt;
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.&lt;br&gt;
I will try to say something about the reasoning I used at each stage of the solution process as I show the solution itself.&lt;br&gt;
The given equation is
{{{sec(x)+tan(x) = 4}}}&lt;br&gt;
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:&lt;br&gt;
{{{1/cos(x)+sin(x)/cos(x) = 4}}}&lt;br&gt;
Next, a general rule in solving equations of almost any type is to clear fractions if possible; so multiply through by cos(x):&lt;br&gt;
{{{1+sin(x) = 4cos(x)}}}&lt;br&gt;
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(x)^2+cos(x)^2 = 1}}}.&lt;br&gt;
With that in mind, let&#39;s square both sides of the equation.  (And remember that, in doing that, we might introduce extraneous solutions to the original problem.)&lt;br&gt;
{{{1+2sin(x)+sin(x)^2 = 16cos(x)^2)}}}&lt;br&gt;
Now use the basic trig identity to get the equation in terms of sin(x) only:&lt;br&gt;
{{{1+2sin(x)+sin(x)^2 = 16(1-sin(x)^2))}}}&lt;br&gt;
Now put this quadratic equation in sin(x) in standard form and factor:&lt;br&gt;
{{{1+2sin(x)+sin(x)^2 = 16-16sin(x)^2)}}}
{{{17sin(x)^2+2sin(x)-15=0}}}
{{{(17sin(x)-15)(sin(x)+1)=0}}}&lt;br&gt;
sin(x) = 15/17  or  sin(x) = -1.&lt;br&gt;
Let&#39;s check these solutions....&lt;br&gt;
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!&lt;br&gt;
If sin(x) = -1, then cos(x) is 0, and both sec(x) and tan(x) are undefined.  So that solution is extraneous.&lt;br&gt;
And now that we have found the solution to the given equation, it is easy to find the answer to the given question:&lt;br&gt;
{{{51sin(x)+34cos(x) = 51*(15/17)+34*(8/17) = 45+16 = 61}}}&lt;br&gt;
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.&lt;br&gt;
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).</PRE>