Number of Integral Solutions for Trigonometric Equation

[utkarshakash]

Homework Statement

Number of integral values of m for the equation $\sin x - \sqrt{3} \cos x = \dfrac{4m-6}{4-m}$ to have a solution is

The Attempt at a Solution

I started by plotting the graph of sinx and √3 cosx. Now if I shift the graph of √3 cosx to √3 units above or below its current position I can still have a solution. In other way 4m-6/4-m can be seen as a numerical constant that determines the shift in the graph of √3 cosx. Imposing the inequality

$- \sqrt{3} \leq \dfrac{4m-6}{4-m} \leq \sqrt{3}$

and solving, I get a different answer.

 

[HallsofIvy]

You might find it easier to use a trig identity to reduce to a single sine.

sin(a+ b)= cos(a)sin(b)+ sin(a)cos(b). Taking b= x, we would like to have cos(a)= 1, $sin(a)= \sqrt{3}$. Of course, that is impossible because $\sqrt{3}$ is larger than 1. But we can "adjust" that. $\sqrt{1^2+ 3^2}= \sqrt{10}$ so $sin(x)- \sqrt{3}cos(x)= \sqrt{10}\left((1/\sqrt{10} sin(x)+ (-\sqrt{3}/\sqrt{10}) cos(x)\right)$. Now we look for a such that $cos(a)= 1/\sqrt{10}$ and $sin(a)= -\sqrt{3}/\sqrt{10}$. That is a= -1.25 radians.

That is, $$sin(3)- \sqrt{3}cos(x)= \sqrt{10}sin(x- 1.25)= \dfrac{4m- 6}{4- m}$$.

Since sine is always between -1 and 1, we must have $$-\sqrt{10}\le\dfrac{4m-6}{4-m}\le \sqrt{10}$$.

 

[Ray Vickson]

You might find it easier to use a trig identity to reduce to a single sine.

sin(a+ b)= cos(a)sin(b)+ sin(a)cos(b). Taking b= x, we would like to have cos(a)= 1, $sin(a)= \sqrt{3}$. Of course, that is impossible because $\sqrt{3}$ is larger than 1. But we can "adjust" that. $\sqrt{1^2+ 3^2}= \sqrt{10}$ so $sin(x)- \sqrt{3}cos(x)= \sqrt{10}\left((1/\sqrt{10} sin(x)+ (-\sqrt{3}/\sqrt{10}) cos(x)\right)$. Now we look for a such that $cos(a)= 1/\sqrt{10}$ and $sin(a)= -\sqrt{3}/\sqrt{10}$. That is a= -1.25 radians.

That is, $$sin(3)- \sqrt{3}cos(x)= \sqrt{10}sin(x- 1.25)= \dfrac{4m- 6}{4- m}$$.

Since sine is always between -1 and 1, we must have $$-\sqrt{10}\le\dfrac{4m-6}{4-m}\le \sqrt{10}$$.

I get $$f(x) = \frac{1}{2} \sin(x - \pi/3).$$ Of course $1^2 + \sqrt{3}^2 = 4$, not 10.

Anyway, could we have not left some of the work for the OP to do?

 

[utkarshakash]

I get $$f(x) = \frac{1}{2} \sin(x - \pi/3).$$ Of course $1^2 + \sqrt{3}^2 = 4$, not 10.

Anyway, could we have not left some of the work for the OP to do?

I think f(x)=2sin(x-∏/3). Doing that way I get the correct answer. But can you please tell what is wrong in my original solution?

 

[Ray Vickson]

I think f(x)=2sin(x-∏/3). Doing that way I get the correct answer. But can you please tell what is wrong in my original solution?

I could not figure out what you were trying to do.