Solving Simple (to Medium-Hard) Trig Equations

Just as with linear equations, I'll first isolate the variable-containing term:

Now I'll use the reference angles I've memorized to get my final answer.

Note: The instructions gave me the interval in terms of degrees, which means that I'm supposed to give my answer in degrees. Yes, the sine, on the first period, takes on the value of 1 at radians, but that's not the angle-measure type they're wanting, and using this as my answer would probably result in my at least losing a few points on this question.

So, in degrees, my answer is:

There's the temptation to quickly recall that the tangent of 60° involves the square root of 3 and slap down an answer, but this equation doesn't actually have a solution. I can see this when I slow down and do the steps. My first step is:

Can any square (of a tangent, or of any other trig function) be negative? No! So my answer is:

The left-hand side of this equation factors. I'm used to doing simple factoring like this:

...and then solving each of the factors. The same sort of thing works here. To solve the equation they've given me, I will start with the factoring:

I've done the algebra; that is, I've done the factoring and then I've solved each of the two factor-related equations. This created two trig equations. So now I can do the trig; namely, solving those two resulting trigonometric equations, using what I've memorized about the cosine wave. From the first equation, I get:

From the second equation, I get:

Putting these the two solution sets together, I get the solution for the original equation as being:

First, I'll get everything over to one side of the "equals" sign:

This equation is "a quadratic in sine"; that is, the form of the equation is the quadratic-equation format:

In the case of the equation they're wanting me to solve, X = sin(θ), a = 1, b = −1, and c = −2.

Since this is quadratic in form, I can apply some quadratic-equation methods. In the case of this equation, I can factor the quadratic:

The first factor gives me the related trig equation:

But the sine is never more than 1, so this equation is not solvable; it has no solution.

The other factor gives me the second related trig equation:

Then my answer is:

I can use a trig identity to get a quadratic in cosine:

The first trig equation, cos(α) = , gives me α = 60° and α = 300°. The second equation gives me α = 180°. So my complete solution is:

I can use a double-angle identity on the right-hand side, and rearrange and simplify; then I'll factor:

The sine wave (from the first trig equation) is zero at 0°, 180°, and 360°. But, in the original exercise, 360° is not included, so this last solution value doesn't count, in this particular instance.

The cosine (from the second trig equation) is at 60°, and thus also at 360° − 60° = 300°. So the complete solution is:

Hmm... I'm really not seeing anything here. It sure would have been nice if one of these trig expressions were squared...

Well, why don't I square both sides, and see what happens?

Huh; go figure: I squared, and got something that I could work with. Nice!

From the last line above, either sine is zero or else cosine is zero, so my solution appears to be:

However (and this is important!), I squared to get this solution, and squaring is an "irreversible" process.

(Why? If you square something, you can't just square-root to get back to what you'd started with, because the squaring may have changed a sign somewhere.)

So, to be sure of my results, I need to check my answers in the original equation, to make sure that I didn't accidentally create solutions that don't actually count. Plugging back in, I see:

It's a good thing that I checked my solutions, because two of them don't actually work. They were created by the process of squaring.

My actual solution is:

2sin(x)cos(x) = 0

2sin(x)cos(x) = sin(2x) = 0