Solving a Trigonometric Equation

In summary, in this conversation, the speaker attempted to solve an equation for all solutions between 0 and 2π, but made a mistake and ended up with an incorrect solution. They provide a summary of the content and suggest that someone else try to solve the equation.
  • #1
FritoTaco
132
23

Homework Statement


Find all solutions of the equation in the interval [itex][0, 2\pi][/itex].

[itex]sin6x+sin2x=0[/itex]

Homework Equations



Double Angle Formulas
[itex]sin2x=2sinxcosx[/itex]

[itex]cos2x=cos^{2}x-sin^{2}x [/itex]
[itex]=2cos^{2}x-1 [/itex]
[itex]=1-2sin^{2}x[/itex]

(3 formulas for cos2x)​

[itex]tan2x=\dfrac{2tanx}{1-tan^{2}x}[/itex]

Sum to Product Formula
[itex]sinA+sinB=2cos\dfrac{A+B}{2}cos\dfrac{A-B}{2}[/itex]

The Attempt at a Solution



[itex]sin6x+sin2x=0[/itex]

[itex]2sin\dfrac{6x+2x}{2}cos\dfrac{6x-2x}{2}=0[/itex] (sum to product formula)

[itex]2sin\dfrac{8x}{2}cos\dfrac{4x}{2}[/itex] (simplify)

[itex]2sin4xcos2x=0[/itex]

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

It became [itex][2(cos^{2}x-1)]^{2}=0[/itex] because 2sin cancels out and the 4 reduces to 2 which is left with, [itex][2(cos^{2}x-1)]^{2}[/itex] 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.
 
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  • #2
FritoTaco said:

Homework Statement


Find all solutions of the equation in the interval [itex][0, 2\pi][/itex].

[itex]sin6x+sin2x=0[/itex]

Homework Equations



Double Angle Formulas
[itex]sin2x=2sinxcosx[/itex]

[itex]cos2x=cos^{2}x-sin^{2}x [/itex]
[itex]=2cos^{2}x-1 [/itex]
[itex]=1-2sin^{2}x[/itex]

(3 formulas for cos2x)​

[itex]tan2x=\dfrac{2tanx}{1-tan^{2}x}[/itex]

Sum to Product Formula
[itex]sinA+sinB=2cos\dfrac{A+B}{2}cos\dfrac{A-B}{2}[/itex]

The Attempt at a Solution



[itex]sin6x+sin2x=0[/itex]

[itex]2sin\dfrac{6x+2x}{2}cos\dfrac{6x-2x}{2}=0[/itex] (sum to product formula)

[itex]2sin\dfrac{8x}{2}cos\dfrac{4x}{2}[/itex] (simplify)

[itex]2sin4xcos2x=0[/itex]

[itex]2sin2xcos2x[2(cos^{2}x-1)]^{2}=0[/itex] (factor)
Mistake here...
FritoTaco said:
It became [itex][2(cos^{2}x-1)]^{2}=0[/itex] because 2sin cancels out
... and here.
FritoTaco said:
and the 4 reduces to 2 which is left with, [itex][2(cos^{2}x-1)]^{2}[/itex] 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.
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.
 
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  • #3
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

Which leads to:

2Sin 2u . Cos u = 0

Can you solve it from here? :-)
 
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  • #4
jmsequeira said:
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

Which leads to:

2Sin 2u . Cos u = 0

Can you solve it from here? :-)
Isn't that what FritoTaco did?
FritoTaco said:
2sin4xcos2x=0
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.
 
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Related to Solving a Trigonometric Equation

Question 1: What is a trigonometric equation?

A trigonometric equation is an equation that involves trigonometric functions such as sine, cosine, tangent, secant, cosecant, and cotangent. These equations often involve one or more unknown angles and require solving for the value of these angles.

Question 2: How do I solve a trigonometric equation?

To solve a trigonometric equation, you need to use the properties and identities of trigonometric functions. This may involve using algebraic manipulation, substitution, or trigonometric identities such as the Pythagorean identity or double angle formulas. It is important to simplify the equation and isolate the unknown angle before solving for its value.

Question 3: What are the common methods for solving trigonometric equations?

Some common methods for solving trigonometric equations include using inverse trigonometric functions, using the unit circle, and using the graphs of trigonometric functions. It is also helpful to remember the properties and identities of trigonometric functions to simplify equations and find solutions.

Question 4: What is the importance of solving trigonometric equations?

Solving trigonometric equations is important in various fields of science and mathematics, such as physics, engineering, and navigation. It allows us to find the values of unknown angles in real-world scenarios and make accurate calculations and predictions.

Question 5: What are some tips for solving trigonometric equations?

Some tips for solving trigonometric equations include checking for extraneous solutions, verifying the solutions by plugging them back into the original equation, and using a calculator to check for approximate solutions. It is also helpful to practice and familiarize yourself with the properties and identities of trigonometric functions.

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