Solving trig equations use both the reference angles and trigonometric identities that you've memorized, together with a lot of the algebra you've learned. Be prepared to need to think in order to solve these equations.
In what follows, it is assumed that you have a good grasp of the trig-ratio values in the first quadrant, how the unit circle works, the relationship between radians and degrees, and what the various trig functions' curves look like, at least on the first period. If you're not sure of yourself, go back and review those topics first.
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Just as with linear equations, I'll first isolate the variable-containing term:
sin(x) + 2 = 3
sin(x) = 1
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:
x = 90°
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:
tan^{2}(θ) = –3
Can any square (of a tangent, or of any other trig function) be negative? No! So my answer is:
no solution
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The left-hand side of this equation factors. I'm used to doing simple factoring like this:
2y^{2} + 3y = 0
y (2y + 3) = 0
...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:
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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:
cos(x) = 0:
x = 90°, 270°
From the second equation, I get:
x = 30°, 330°
Putting these the two solution sets together, I get the solution for the original equation as being:
x = 30°, 90°, 270°, 330°
First, I'll get everything over to one side of the "equals" sign:
sin^{2}(θ) – sin(θ) – 2 = 0
This equation is "a quadratic in sine"; that is, the form of the equation is the quadratic-equation format:
aX^{2} + bX + c = 0
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:
sin^{2}(θ) – sin(θ) – 2 = 0
(sin(θ) – 2)(sin(θ) + 1) = 0
The first factor gives me the related trig equation:
sin(θ) = 2
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:
sin(θ) + 1 = 0
sin(θ) = –1
Then my answer is:
(If you're doing degrees-only solutions in your class, the solution value above equates to "270°".)
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I can use a trig identity to get a quadratic in cosine:
cos^{2}(α) + cos(α) = sin^{2}(α)
cos^{2}(α) + cos(α) = 1 – cos^{2}(α)
2cos^{2}(α) + cos(α) – 1 = 0
(2cos(α) – 1)(cos(α) + 1) = 0
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:
α = 60°, 180°, 300°
I can use a double-angle identity on the right-hand side, and rearrange and simplify; then I'll factor:
sin(β) = 2sin(β)cos(β)
sin(β) – 2sin(β) cos(β) = 0
sin(β)(1 – 2cos(β)) = 0
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:
β = 0°, 60°, 180°, 300°
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?
(sin(x) + cos(x))^{2} = (1)^{2}
sin^{2}(x) + 2sin(x)cos(x) + cos^{2}(x) = 1
[sin^{2}(x + cos^{2}(x)] + 2sin(x)cos(x) = 1
1 + 2sin(x)cos(x) = 1
2sin(x)cos(x) = 0
sin(x)cos(x) = 0
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:
x = 0°, 90°, 180°, 270°
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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:
sin(0°) + cos(0°) = 0 + 1 = 1
...so the solution "x = 0°" works
sin(90°) + cos(90°) = 1 + 0 = 1
...so the solution "x = 90°" works, too
sin(180°) + cos(180°) = 0 + (–1) = –1
...oh, okay, so "x = 180°" does NOT work
sin(270°) + cos(270°) = (–1) + 0 = –1
...so "x = 270°" doesn't work, either
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:
x = 0°, 90°
Note: In the above, I could have stopped at this line:
2sin(x)cos(x) = 0
...and used the double-angle identity for sine, in reverse, instead of dividing off the 2 in the next-to-last line in my computations. The answer would have been the same, but I would have needed to account for the solution interval:
2sin(x)cos(x) = sin(2x) = 0
Then 2x = 0°, 180°, 360°, 540°, etc, and dividing off the 2 from the x would give me x = 0°, 90°, 180°, 270°, which is the same almost-solution as before. After doing the necessary check (because of the squaring) and discarding the extraneous solutions, my final answer would have been the same as previously.
The squaring trick in the last example above doesn't come up often, but if nothing else is working, it might be worth a try. Keep it in mind for the next test.
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