Solvin trig equation

  • Thread starter mohlam12
  • Start date
  • #1
154
0

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
mathman
Science Advisor
7,759
415
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
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
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
154
0
yes, that s interestin'
 
  • #5
154
0
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
VietDao29
Homework Helper
1,423
1
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
154
0
hmmm yeah!
thank you!

Mohammmed
 

Related Threads for: Solvin trig equation

  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
13
Views
5K
Replies
3
Views
4K
  • Last Post
Replies
5
Views
3K
Replies
3
Views
3K
  • Last Post
Replies
10
Views
3K
Top