Solving Trigonometric Equations (page 1 of 2)
Solving trig equations use both the reference angles you've memorized and a lot of the algebra you've learned. Be prepared to need to think!
Just as with linear equations, I'll first isolate the variable-containing term:
+ 2 = 3
Now I'll use the reference angles I've memorized:
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:
tan2(x) = –3
How can the square of a trig function evaluate to a negative number? It can't!
To solve this, I need to do some simple factoring:
Now that I've done the algebra, I can do the trig. From the first factor, I get x = 90° and x = 270°. From the second factor, I get x = 30° and x = 330°.
x = 30°, 90°, 270°, 330° Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved
This is a quadratic in sine, so I can apply some of the same methods:
– sin(x) – 2 = 0
Only one of the factor solutions is sensible. For sin(x) = –1, I get:
x = 270°
I can use a trig identity to get a quadratic in cosine:
+ cos(x) = sin2(x)
The first trig equation, cos(x) = 1/2, gives me x = 60° and x = 300°. The second equation gives me x = 180°. So my complete solution is:
x = 60°, 180°, 300°
I can use a double-angle identity on the right-hand side, and rearrange and simplify; then I'll factor:
I can The sine wave is zero at 0°, 180°, and 360°. (But: In the original exercise, 360° is not included, so this last solution doesn't count, in this particular instance.) The cosine is 1/2 at 60°, and thus also at 360° – 60° = 300°. So the complete solution is:
x = 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, then, and see what happens?
cos(x))2 = (1)2
Huh; go figger: 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°
However (and this is important!), I squared to get this solution, so 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:
= 0 + 1 = 1 (this
So the actual solution is:
x = 0°, 90°
Note that I could have 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 before.
This squaring trick doesn't come up often, but if nothing else is working, it might be worth a try.