Solving Harder Trig Equations
Purplemath
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Solve sin(x/2) = cos(x/2) in full generality.
There are various ways of going about this, but I think I'll take an easy way out. By dividing through by the cosine, I'll get a tangent:
(Why is this division okay? Well, I can't divide by zero, so this division is only okay if cosine isn't equal to zero. But the original equation had sine and cosine equal, and they're never zero at the same place. And the tangent is never equal to one where the cosine is equal to zero. So there was no division-by-zero issue, in this case. But always remember to check yourself, to be sure.)
The tangent is equal to 1 for on the first period. But this exercise wants the answer "in full generality". Obviously, I can't list out all of the solution values, because there are infinitely many of them. So I'll have to use a formula.
From what I know about the graph of the tangent, I know that the tangent will equal 1 at 45° after every 180°. These solutions for are at 0° + 45°, 180° + 45°, 360° + 45°, and so forth. To give the answer "in full generality", I'll use a formula:
, for all integers n
Now I need to solve for x itself. I'll multiply through by 2:
x = (360n)° + 90°
In radians, the solution above would be etc; and the general solution would be
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Solve 3tan3(x) − 3tan2(x) − tan(x) + 1 = 0 in full generality.
I can factor this in pairs:
3tan2(x)[tan(x) − 1] − 1[tan(x) − 1] = 0
[tan(x) − 1][3tan2(x) − 1] = 0
The first equation solves, in the first period, as:
x = 45°, 225°
The second solves, in the first period, as:
x = 30°, 150°, 210°, 330°
To make the solution "general", I need to state the above solutions formulaically, to account for every period.
The first solution is 45° more than a multiple of 180°, so (180n)° + 45° should do. The second solution is 30° more than a multiple of 180° and (because of the "plus / minus") also 30° less than that same multiple, so (180n)° ± 30° will cover this part.
x = (180n)° ± 30°, (180n)° + 45° for all integers n
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Solve
![2sin(x) − 2sqrt[3]cos(x) − sqrt[3]tan(x) + 3 = 0](trig/slvtrg03.gif)
on [0, 2π)
What on earth...?!?
When nothing looks like it's going to work, sometimes it helps to put everything in terms of sine and cosine. That process, applied to this equation, gives me:
![2sin(x) − 2sqrt[3]cos(x) − sqrt[3]sin(x)/cos(x) + 3 = 0](trig/slvtrg05.gif){kind=small}
That's not a whole lot better... but the first two terms share a common factor of 2. If I convert the last term to a common denominator with the third term, what will that give me?
![2sin(x) − 2sqrt[3]cos(x) − sqrt[3]sin(x)/cos(x) + 3cos(x)/cos(x) = 0](trig/slvtrg06.gif){kind=small}
If I factor a 2 from the first two terms and the square root of 3 and a cosine from the second two terms, I'll get:
![2(sin(x) − sqrt[3]cos(x)) − (sqrt[3]/cos(x))(sin(x) − sqrt[3]cos(x)) = 0](trig/slvtrg07.gif){kind=small}
Now I can take the common factor out front:
![(sin(x) − sqrt[3]cos(x))(2 − sqrt[3]/cos(x)) = 0](trig/slvtrg08.gif){kind=small}
Whew! That actually worked! Okay, now I need to solve the factors. The first factor solves as:
![sin(x) = sqrt[3]cos(x), tan(x) = sqrt[3]](trig/slvtrg09.gif)
This equation is true at x = 60° and, by the symmetry of the tangent curve, also at x = 180° + 60° = 240°. In radians, this is 
.
The second factor solves as:
![2 = sqrt[3]/cos(x), cos(x) = sqrt[3]/2](trig/slvtrg11.gif)
Cosine takes on this value at x = 30° and, by the symmetry of the cosine curve, also at x = 360° − 30° = 330°. In radians, this is 
. So my solution is:
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Solve ln(2 − sin2(x)) = 0 on 0° < x < 360°
The natural log (well, any log) is zero when the argument is 1, so this gives me:
2 − sin2(x) = 1
1 − sin2(x) = 0
(1 − sin(x))(1 + sin(x)) = 0
1 = sin(x) or 1 = −sin(x)
From what I know of the sine wave, my solution is:
x = 90°, 270°
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Solve on [0, 2π)
By nature of logarithms, the equivalent exponential equation is:
![3^(1/2) = sqrt[3] = 2sin(x), sqrt[3]/2 = sin(x)](trig/slvtrg14.gif)
The sine takes on this value at 
and also at 
. Then my solution is:
Expect to need to factor (especially quadratics) in order to solve some trig equations, and also expect to need to use trig identities. Don't be afraid to try different methods; sometimes your first impulse doesn't lead anywhere helpful, but your second guess might work fine. And pay particular attention to any oddly complex examples in your textbook, as these may hold hints about what tricks you will need, especially on the next test.
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