# Trigonometric Equations

• September 1st 2008, 03:29 AM
magentarita
Trigonometric Equations
Solve each trigonometric equation on the interval
0 <or=to x < 2pi....Does this mean to find all the answers between 0 degrees and 360 degrees?

(1) cos^2 (x) - sin^2 (x) + sin (x) = 0

My attempt:

cos^2 (x) = 1 - sin^2 (x)

I replace cos^2 (x) with 1 - sin^2 (x) in the equation, right?

I get this:

1 - sin^2 (x) - sin^2 (x) + sin (x) = 0

Do I factor out sin(x)? If so, what happens next to get me on the right track to the answer? OR....do I rearrange the sin(x) terms to be in the form ax^2 + bx + c?

(2) sin(4x) - sin(6x) = 0

I know there is a sin(2x) but I have never seen a 4x or 6x applied to a trig function.

Does sin(4x) = sin(2x) + sin(2x)?

Does sin(6x) = sin(3x) + sin(3x)?

If so, where do I go from there?

• September 1st 2008, 03:35 AM
mr fantastic
Quote:

Originally Posted by magentarita
Solve each trigonometric equation on the interval
0 <or=to x < 2pi....Does this mean to find all the answers between 0 degrees and 360 degrees?

(1) cos^2 (x) - sin^2 (x) + sin (x) = 0

My attempt:

cos^2 (x) = 1 - sin^2 (x)

I replace cos^2 (x) with 1 - sin^2 (x) in the equation, right?

I get this:

1 - sin^2 (x) - sin^2 (x) + sin (x) = 0

Do I factor out sin(x)? If so, what happens next to get me on the right track to the answer? OR....do I rearrange the sin(x) terms to be in the form ax^2 + bx + c?

[snip]

$1 - \sin^2 x - \sin^2 x + \sin x = 0$

$\Rightarrow 2 \sin^2 x - \sin x - 1 = 0$

$\Rightarrow (2 \sin x + 1)(- \sin x + 1) = 0$

etc.
• September 1st 2008, 03:40 AM
mr fantastic
Quote:

Originally Posted by magentarita
[snip]
(2) sin(4x) - sin(6x) = 0

I know there is a sin(2x) but I have never seen a 4x or 6x applied to a trig function.

Does sin(4x) = sin(2x) + sin(2x)?

Does sin(6x) = sin(3x) + sin(3x)?

If so, where do I go from there?

General solutions:

$\sin (4x) = \sin (6x)$

Case 1: $4x = 6x + 2 n \pi$ where n is an integer.

Case 2: $4x = (\pi - 6x) + 2 n \pi$ where n is an integer.

In each case use the general solution to solve for x over the given domain.
• September 1st 2008, 07:45 AM
Soroban
Hello, magentarita!

Quote:

Solve on the interval $0 \leq x < 2\pi$

$(2)\;\;\sin(4x) - \sin(6x) \:= \:0$

Does $\sin(4x) \:= \:\sin(2x) + \sin(2x)$ ? . . . . sorry, no
Sum-to-Product identity: . $\sin A - \sin B \;=\;2\cdot\cos\!\left(\frac{A+B}{2}\right)\cdot\s in\!\left(\frac{A-B}{2}\right)$

We have: . $\sin(4x) - \sin(6x) \;=\;0$

. . $2\cdot\cos\!\left(\frac{4x+6x}{2}\right)\cdot\sin\ !\left(\frac{4x-6x}{2}\right)\;=\;0$

. . . . . . $2\cdot\cos(5x)\cdot\sin(\text{-}x) \;=\;0$

. . . . . . $\text{-}2\cdot\cos(5x)\cdot\sin(x) \;=\;0$

We have: . $\sin(x) \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:0,\:\pi}$

And: . $\cos(5x) \:=\:0\quad\Rightarrow\quad 5x \:=\:\frac{\pi}{2} + \pi n \quad\Rightarrow\quad x \:=\:\frac{\pi}{10} + \frac{\pi}{5}n$

. . Hence: . $\boxed{x \;=\;\frac{\pi}{10},\;\frac{3\pi}{10},\;\frac{\pi} {2},\;\frac{7\pi}{10},\;\frac{9\pi}{10},\;\frac{11 \pi}{10},\;\frac{13\pi}{10},\;\frac{3\pi}{2},\;\fr ac{17\pi}{10},\:\frac{19\pi}{10}}$

This results in the same answers as in Mr. F's excellent/elegant solution.
• September 1st 2008, 09:09 PM
magentarita
Thanks
I thank you both, especially Soroban for making life simple for students like me.
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