# Trigonometric Equations that Require Factoring - Problem 3

We are solving trigonometric equations that require factoring. Now this one is a slightly harder example because you’ll notice that we’ve got sines and cosines all mixed together here. Now usually what you want to do is you want to make sure you equation has just one of the other. Unless it's obvious how to factor and here it isn’t.

So the first thing I would do is I would replace sine squared with one minus cosine squared in order to get everything in terms of cosine. So I’ll have minus 2 times 1 minus cosine squared. Minus 5 cosine x plus 4 equals 0. And then I want to distribute the -2 through and simplify. So -2 times 1, -2, -2 times negative cosine squared is plus 2 cosine squared x minus 5 cosine x plus 4 equals 0. And then the minus 2 and the 4 are going to give me just plus 2.

So I have 2 cosine squared x minus 5 cosine x, plus 2 equals 0. At this point we want to try to factor now the 2 cosine squared x suggests that this should start out as 2 cosine x and cosine x. And then the two at the end so just say I need a 1 and a 2. The question is how do I set this up? I want to get 5 in the middle, so maybe I should have a 2 cosine x being multiplied by 2 and this cosine being multiplied 1, that will give me 5. So if I make both of these negative I’ll get negative cosine minus 4 cosine minus 5 in the middle of the way I need.

Now that we’ve done that, we have to figure out what kind of basic trig equations we are solving here. Now in order for this to be 0 cosine x has to equal ½ or cosine x equals 2. When you see something like this I want you to stop immediately, that should definitely wring some alarm bells.

Cosine x equals 2 is impossible, the cosine function looks like this. It goes between negative 1 and 1 it never gets up to 2. So this has no solutions just this little part here, we will get some solutions from here. Cosine x equals a half, on the unit circle that will be represented by points will give me first coordinate at a half second coordinate at something else.

So let me just it’s another point down here,1/2,-1. What angle will this be? You really should memorize angles, the special angle on the first quadrant at least so that you know that this is can be pi over 3. Cosine of pi over 3 is one half. And that gives you one solution but you should also memorize cosines and even functions so opposite inputs have the same output and that’s why negative pi over 3 is also a solution because it gives you a cosine of one half. So plus or minus pi over 3.

Now to get the rest of the solutions you need to use periodicity. So x equals plus or minus pi over 3 plus 2n pi, right 2 times any integer times 5. Now for example if you took this angle pi over 3 and added 2pi you would still be at an angle that gave you a cosine of a half. So this is all our solutions because the second equation that we got didn’t have any solutions watch of for this. Cosine x equals 2 cosine x equals 5 cosine x equals minus 1 point 3, these are all of the equations that have no solutions.

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