Solving Trig Equations - Get Expert Help Now!

[mohlam12]

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!
if someonce can tell me please how to solve equations like that! ( -infinity < x < infinity)

thank you again,
mohamed

 

[mathman]

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.

 

[arildno]

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!
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:
\cos(u-v)=\cos(u)\cos(v)+\sin(u)\sin(v) (1)

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?

 

[mohlam12]

yes, that s interestin'

 

[mohlam12]

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 ?!

 

[VietDao29]

So:
\cos(2x) + 3 \sin(2x) - 1 = 0
You then divide both sides by : \sqrt{1 + 3 ^ 2} = \sqrt{10}. So, you have:
\frac{1}{\sqrt{10}} \cos(2x) + \frac{3}{\sqrt{10}} \sin(2x) = \frac{1}{\sqrt{10}}
Now let:
\cos \alpha = \frac{1}{\sqrt{10}}, and \sin \alpha = \frac{3}{\sqrt{10}}. You can do this because you know that:
\left( \frac{1}{\sqrt{10}} \right) ^ 2 + \left( \frac{3}{\sqrt{10}} \right) ^ 2 = 1, so it's true that \cos ^ 2 \alpha + \sin ^ 2 \alpha = 1
So you have:
\cos \alpha \cos(2x) + \sin \alpha \sin(2x) = \cos \alpha
\Leftrightarrow \cos(\alpha - 2x) = \cos \alpha
Can you go from here?
-------------------------------
Or you can do it a little bit differently:
\cos (2x) + 3 \sin (2x) - 1 = 0
\Leftrightarrow 1 - 2 \sin ^ 2 x + 6 \sin x \cos x - 1 = 0
\Leftrightarrow - 2 \sin ^ 2 x + 6 \sin x \cos x = 0
\Leftrightarrow - \sin ^ 2 x + 3 \sin x \cos x = 0
\Leftrightarrow \sin x (3 \cos x - \sin x) = 0
Can you go from here?
Viet Dao,

 

[mohlam12]

hmmm yeah!
thank you!

Mohammmed