Solving Trig Equations - Get Expert Help Now!

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In summary: The solution involves using the trigonometric relation for cosine of a sum or difference and transforming the equation into a form similar to Acos(X)+B=0. This can be done by letting a=fcos(y) and b=fsin(y), where f=sqrt(a^2+b^2) and tan(y)=b/a. Then, using the sum formula, the equation can be simplified to solve for the variable x. Alternatively, the equation can also be solved by factoring and setting each factor equal to 0.
  • #1
mohlam12
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Hi,
I just learned at class how to solve trig equations in this form:
acos(x)+bsin(x)+c=0

but i didnt understand ANYTHING!:eek:
if someonce can tell me please how to solve equations like that! ( -infinity < x < infinity)

thank you again,
mohamed
 
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  • #2
One simple trick is to let a=fcos(y) and b=fsin(y), where f=sqrt(a^2+b^2) and tan(y)=b/a. Then fcos(x-y)+c=0. You can go from there. Notice that if |c|>|f|, there is no real solution.
 
  • #3
mohlam12 said:
Hi,
I just learned at class how to solve trig equations in this form:
acos(x)+bsin(x)+c=0

but i didnt understand ANYTHING!:eek:
if someonce can tell me please how to solve equations like that! ( -infinity < x < infinity)

thank you again,
mohamed
Hi, the first thing you should focus on understanding, is the trigonometric relation between the cosine (or sine) of a sum (or difference), and the cosines and sines of the terms of the sum, that is, for example:
[tex]\cos(u-v)=\cos(u)\cos(v)+\sin(u)\sin(v) (1)[/tex]

Note that the "u"-variable appears in two terms on your right-hand side, in the cos(u) term and the sin(u) term.

Suppose you had an trigonometric equation of the following form:
Acos(X)+B=0(2)
Would you find that difficult to solve?
Probably not!

Go now back to your given equation: acos(x)+bsin(x)+c=0
What if we could utilize the sum formula (1) in a smart manner so that we could transform our equation into something like (2)?
This is the rationale behind our solution strategy!

Do you follow thus far?
 
  • #4
yes, that s interestin'
 
  • #5
so yeah, let's take this example:
cos(2x) + 3sin(2x) - 1 = 0
so we have f=sqrt(10) and then sqrt(10)cos(x-y)-1=0
where can i go from here ?!
 
  • #6
So:
[tex]\cos(2x) + 3 \sin(2x) - 1 = 0[/tex]
You then divide both sides by : [itex]\sqrt{1 + 3 ^ 2} = \sqrt{10}[/itex]. So, you have:
[tex]\frac{1}{\sqrt{10}} \cos(2x) + \frac{3}{\sqrt{10}} \sin(2x) = \frac{1}{\sqrt{10}}[/tex]
Now let:
[tex]\cos \alpha = \frac{1}{\sqrt{10}}[/tex], and [tex]\sin \alpha = \frac{3}{\sqrt{10}}[/tex]. You can do this because you know that:
[tex]\left( \frac{1}{\sqrt{10}} \right) ^ 2 + \left( \frac{3}{\sqrt{10}} \right) ^ 2 = 1[/tex], so it's true that [tex]\cos ^ 2 \alpha + \sin ^ 2 \alpha = 1[/tex]
So you have:
[tex]\cos \alpha \cos(2x) + \sin \alpha \sin(2x) = \cos \alpha[/tex]
[tex]\Leftrightarrow \cos(\alpha - 2x) = \cos \alpha[/tex]
Can you go from here?
-------------------------------
Or you can do it a little bit differently:
[tex]\cos (2x) + 3 \sin (2x) - 1 = 0[/tex]
[tex]\Leftrightarrow 1 - 2 \sin ^ 2 x + 6 \sin x \cos x - 1 = 0[/tex]
[tex]\Leftrightarrow - 2 \sin ^ 2 x + 6 \sin x \cos x = 0[/tex]
[tex]\Leftrightarrow - \sin ^ 2 x + 3 \sin x \cos x = 0[/tex]
[tex]\Leftrightarrow \sin x (3 \cos x - \sin x) = 0[/tex]
Can you go from here?
Viet Dao,
 
Last edited:
  • #7
hmmm yeah!
thank you!

Mohammmed
 

FAQ: Solving Trig Equations - Get Expert Help Now!

What is a trigonometric equation?

A trigonometric equation is an equation that contains trigonometric functions, such as sine, cosine, tangent, etc. The goal is to find the values of the variables that make the equation true.

What is the process for solving a trigonometric equation?

The process for solving a trigonometric equation involves using algebraic manipulation and trigonometric identities to isolate the variable and solve for its value. This may also involve using inverse trigonometric functions and graphing techniques.

What are the common techniques used for solving trigonometric equations?

Some common techniques for solving trigonometric equations include factoring, using the unit circle, using trigonometric identities, and using algebraic manipulation. It is important to choose the most appropriate technique depending on the type of equation and the given information.

Why is it important to check the solutions of a trigonometric equation?

It is important to check the solutions of a trigonometric equation because there may be extraneous solutions, which are solutions that do not satisfy the original equation. Checking the solutions helps to ensure that the final answer is correct.

How do I get expert help for solving trigonometric equations?

You can get expert help by consulting a math tutor or attending a math tutoring session. Additionally, there are many online resources and forums where you can ask for help from knowledgeable individuals. It is important to seek help from reliable sources to ensure the accuracy of the solutions.

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