# Solving a Trigonometric Equation

1. Dec 1, 2016

### FritoTaco

1. The problem statement, all variables and given/known data
Find all solutions of the equation in the interval $[0, 2\pi]$.

$sin6x+sin2x=0$

2. Relevant equations

Double Angle Formulas
$sin2x=2sinxcosx$

$cos2x=cos^{2}x-sin^{2}x$
$=2cos^{2}x-1$
$=1-2sin^{2}x$

(3 formulas for cos2x)​

$tan2x=\dfrac{2tanx}{1-tan^{2}x}$

Sum to Product Formula
$sinA+sinB=2cos\dfrac{A+B}{2}cos\dfrac{A-B}{2}$

3. The attempt at a solution

$sin6x+sin2x=0$

$2sin\dfrac{6x+2x}{2}cos\dfrac{6x-2x}{2}=0$ (sum to product formula)

$2sin\dfrac{8x}{2}cos\dfrac{4x}{2}$ (simplify)

$2sin4xcos2x=0$

$2sin2xcos2x[2(cos^{2}x-1)]^{2}=0$ (factor)

It became $[2(cos^{2}x-1)]^{2}=0$ because 2sin cancels out and the 4 reduces to 2 which is left with, $[2(cos^{2}x-1)]^{2}$ and it's squared because the cos2x is used in the Double Angle Formula. I don't know where to go from here, any suggestions? Thanks.

2. Dec 1, 2016

### Staff: Mentor

Mistake here...
... and here.
In the first mistake, $\cos(2x) = 2\cos^2(x) - 1$, which is different from what you wrote.
In the second mistake, "cancelling" is not a viable option, as you lose solutions. Here's a simple example:
x(x - 1) = 0
"Cancel" x to get x - 1 = 0, or x = 1
This is incorrect, or at least incomplete, as x = 0 is a solution of the original equation.

3. Dec 1, 2016

### jmsequeira

You could try:

2x = u or x = u/2

Then you get:

sin u + sin 3u = 0

If we use "sum to product formula" you get:

A+B = 3u + u = 4u
A-B = 2u

2Sin 2u . Cos u = 0

Can you solve it from here? :-)

4. Dec 3, 2016

### haruspex

Isn't that what FritoTaco did?
From here, you could have proceeded
sin(2x)cos2(2x)=0
sin(2x)cos(2x)=0
Cancelling out one cos(2x) factor is ok because we have one left, in case that is the zero factor
sin(4x)=0
etc.