How do you solve acos(x)+bsin(x)+c=0?

[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