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Problem

Consider the trigonometric equation: $$ a\sin x+b\cos x-\cos x\sin x=0\qquad(0\le x<2\pi)\tag{*} $$ try to analyze the number of solutions to equation (*) with parameters $a,b$, i.e, let $A=a^{2/3}+b^{2/3}-1$, we have:

  1. $A<0$, there are four distinct solutions.
  2. $A>0$, there are two distinct solutions.

Endeavors

Let $f(x)=a\sin x+b\cos x-\cos x\sin x$, we have $f^\prime(x)=a\cos x-b\sin x-\cos2x$. It seems no advance to calculate the derivative, because $f^\prime$ is as hard as $f$.

Let $u=\cos x$ and $v=\sin x$, we have $u^2+v^2=1$ and $av+bu=uv$. We can work on these equations, but I prefer the trigonometric way, i.e, analyze the properties of $f(x)$.

I want to illustrate some details about $f(x)$, which might be useful. Let $a=r\cos\phi$ and $b=r\sin\phi$, where $r=\sqrt{a^2+b^2}$, we have $f(x)=r\sin(x+\phi)-\frac12\sin2x$. It's a linear combination of $\sin(x+\phi)$ and $\sin2x$. I don't know whether there's a systematical way to deal with it.

Any idea? Thanks!

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I think you need to be clear what you mean by distinct solutions. For example, if we set $a=0$ we find that $\sin x = b \text{ or } \cos x = 0$ - what count as distinct solutions of these equations? –  Mark Bennet Jan 4 '13 at 9:45
    
$|a\sin(x)+b\cos(x)|\leq\sqrt{a^2+b^2}$ –  Babak S. Jan 4 '13 at 9:53
    
@MarkBennet Sorry, $0\le x<2\pi$. –  Frank Science Jan 4 '13 at 12:20
    
@FrankScience. For $a=0$ and $b=1$ you have $A=0$ and there are one double root and one simple root for the function $f(x)=a*\sinx +b*\cosx-\cosx\sinx$. The same for $a=0$ and $b=-1$ and for $a=0$ and $b=\pm1$. Why it's said two double solutions? –  RicardoCruz Jan 5 '13 at 10:49
    
@RicardoCruz Maybe the concept of double root in transcendental equation is ambiguous, so I've edited. Thanks. –  Frank Science Jan 5 '13 at 15:11

1 Answer 1

up vote 2 down vote accepted

Let us consider only four cases in our analysis. The other missing cases can be analyzed by a similar approach.

Let $A$ a real number such that $A=a^\frac{2}{3}+b^\frac{2}{3}-1$ and $g(x)= a\sin x + b\cos x-\cos x\sin x$ a function with domain $[0, 2\pi[$ , let's analyze the number of zeros of $g(x)$ in function of $A$.

(1) Case ($a=0$ and $b \neq 0$)

$$g(x)=0 \Leftrightarrow b\cos x-\cos x\sin x = 0$$

$$ \Leftrightarrow \cos x(b-\sin x)=0$$

For $b^2>1$ we have $A>0$ and two distinct solutions.

For $b^2<1$ we have $A<0$ and four distinct solutions.

For $b^2=1$ we have $A=0$ and a double root and a simple one.

(2) Case ($a \neq 0$ and $b = 0$)

$$g(x)=0 \Leftrightarrow a\sin x-\cos x\sin x=0$$ $$\Leftrightarrow \sin x(a-\cos x)=0$$

For $a^2>1$ we have $A>0$ and two distinct solutions.

For $a^2<1$ we have $A<0$ and four distinct solutions.

For $a^2=1$ we have $A=0$ and a double root and a simple one.

(3) Case ($a=0$ and $b = 0$) $$g(x)=0 \Leftrightarrow -\cos x\sin x=0$$ We have $A<0$ and the equation has four distinct solutions.

(4) Case ($a>0$ and $b>0$) $$g(x)=0 \Leftrightarrow a\sin x + b\cos x-\cos x \sin x=0$$ For the equation above $0$, $\frac{\pi}{2}$, $\pi$, and $\frac{3\pi}{2}$ are never solutions (Verify by yourself). Let's make the following factorising: $$g(x)=(a\tan x + b - \sin x)\cos x$$ And let's define $f(x)=a\tan x + b - \sin x$. Every root of $f(x)$ is a root of $g(x)$. So let's search for roots of $f(x)$. $$f(x)=0 \Leftrightarrow a\tan x + b= \sin x$$ If we change the values of $a$ and $b$ ($a>0$ and $b>0$) the graph of $a\tan x+b$ will always cross the graph of $\sin x$ once in the intervals $[\frac{\pi}{2}, \frac{3\pi}{2}]$ and $[\frac{3\pi}{2}, 2\pi[$. So $f(x)$ has at least two distinct roots in $[0, 2\pi[$.

See the graph below:

Atanx+bAndsinx

For $x \in [0, \frac{\pi}{2}]$ we must analyze the values of $a$ and $b$. The local minimum of $f(x)$ in $[0, \frac{\pi}{2}]$ occurs at $\cos x_0 = a^\frac{1}{3}$ i.e. $x_0= \arccos a^\frac{1}{3}$. If we substitute $x_0$ in $f(x)$ we get: $$f(x_0)= \frac{a \sin (\arccos a^\frac{1}{3})}{a^\frac{1}{3}} + b - \sin (\arccos a^\frac{1}{3})$$ After some algebra we get: $$f(x_0) = -(1-a^\frac{2}{3})^\frac{3}{2} + b$$ Now let's analyze the possibilities. Note that

$$ f(x_0) >0 \Leftrightarrow -(1-a^\frac{2}{3})^\frac{3}{2} + b>0 \Leftrightarrow b>(1-a^\frac{2}{3})^\frac{3}{2}$$ $$\Leftrightarrow a^\frac{2}{3}+ b^\frac{2}{3} -1 >0 \Leftrightarrow A >0.$$ So

If $f(x_0) > 0$ ($A>0$), then $a\tan x +b$ and $\sin x$ have no intersection ($f(x_0) >0$), $f(x)$ has only two distinct roots in $[0, 2\pi[$ and so do $g(x)$.

If $f(x_0) =0$ ($A = 0$), then $a\tan x +b$ and $\sin x$ are tangent at $x_0$, $f(x)$ has a double root in $[0,\frac{\pi}{2}]$ (a double root and two distinct ones in $[0, 2\pi[$) and so do $g(x)$.

If $f(x_0)<0$ ($ A < 0)$, then $a\tan x +b$ and $\sin x$ have two intersections, $f(x)$ has only two distinct roots in $[0,\frac{\pi}{2}]$ (four distinct roots in $[0, 2\pi[$) and so do $g(x)$.

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Pretty good. Thanks. –  Frank Science Jan 6 '13 at 5:31
    
@FrankScience. You're wellcome:) –  RicardoCruz Jan 6 '13 at 10:08

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