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Solving Trig Equations: Examples (page 2 of 2)

• 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:

tan(x/2) = 1

(This division was okay, because I know from the graphs that the cosine isn't equal to zero anywhere that it's equal to the sine of the same angle.)

The tangent is equal to 1 for x/2 = 45° and 225° 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. 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 tan(x/2) are at 0° + 45°, 180° + 45°, 360° + 45°, and so forth. To give the answer "in full generality", I'll use a formula:

x/2 = (180×n)° + 45°, for all integers n

Now I need to solve for x itself. I'll multiply through by 2:

x = (360n)° + 90°

• 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
tan(x) = 1  or  tan²(x) = 1/3
tan(x) = 1  or  tan(x) = ±1 / sqrt[3]

The first equation solves as x = 45° and 225°; the second solves as x = 30°, 150°, 210°, and 330°. To make the solution "general", I need to state these formulaically.

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 the same multiple, so (180n)° ± 30° will cover this part.

x = (180n)° ± 30°, (180n)° + 45° for all integers n

• Solve on

What on earth...?

When nothing works, sometimes it helps to put everything in terms of sine and cosine. That process, applied to this equation, gives me:

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?

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:

Now I can take the common factor out front:

Whew! That actually worked! Okay, now I need to solve the factors. The first factor solves as:

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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 . And the second factor solves as:

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:

• Solve ln(2 – sin²(x)) = 0 on 0° < x < 360°

The natural 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) 
1 = –sin(x) 

From what I know of the sine wave, my solution is:

x = 90°, 270°

• Solve log₃(2sin(x)) = 1/2 on [0, 2π)

By nature of logarithms, the equivalent exponential equation is:

The sine takes on this value at and also at . Then my solution is:

Expect to need to factor (especially quadratics) and 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 test.

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