# Equation for Intersection Points of Graphs

• Jun 7th 2011, 04:57 AM
starshine84
Equation for Intersection Points of Graphs
The graphs of f(x) = 2sin(x) - 1 (blue) , and g(x) = 3cos(x) + 2 (red)
are shown below:

https://www.virtualhighschool.com/co.../images/38.jpg

What equation would have the intersection points of the graphs as its solutions?

I know that I need to set the equations equal to one another:

2sin(x) -1 = 3cos(x) + 2
2sin(x) -1 -2 = 3cos(x)
2sin(x) -3 = 3cos(x)

I think I need to get all of the trig functions in terms of the same ratio, but I'm not sure how to do that. Nor am I sure where to go from there.

Any help would be greatly appreciated. This is my last trig question and I'm finally done the unit!

Thanks :)
• Jun 7th 2011, 05:37 AM
starshine84
Okay, I think I have gotten a little bit further, but I'm still stuck....

I have done the following:

2sin(x) -3 = 3cos(x)
2sin(x) -3 = 3 /sqrt [1-sin^2(x)]
[2sin(x) -3][2sin(x) -3] = 9 [1-sin^2(x)]
4sin^2(x) -6sin(x) - 6sin(x) + 9 = 9 - 9sin^2(x)
13sin^2(x) - 12sin(x) = 0
sin(x) [13sin(x) - 12] = 0

sin(x) = 0
Quadrants- I, II, III, and IV
Reference Angles= 0, 180, 360
x= 0, 180, 360

13sin(x) - 12 = 0
13sin(x) = 12
sin(x) = 12/13
x = 67.38
Reference angle= 67.38

0+67.38 = 67.38
180-67.38 = 112.62

x= 67.38 and 112.62

Therefore, x= 1, 67.38, 112.62, 180, and 360

But those values don't answer the question. The question asks to find what equation would have the intersection points of the graphs as its solutions?

So, I'm still lost....
• Jun 7th 2011, 07:06 AM
Soroban
Hello, starshine84!

Quote:

$\displaystyle \text{Consider the graphs of: }\,f(x) \:=\: 2\sin x - 1\,\text{ and }\,g(x) \:=\: 3\cos x + 2$

$\displaystyle \text{What equation would have the intersection points of the graphs as its solutions?}$

$\displaystyle \text{I know that I need to set the equations equal to one another:}$ . Yes!

. . $\displaystyle 2\sin x -1 \:=\: 3\cos x + 2 \quad\Rightarrow\quad 2\sin x -3 \:=\: 3\cos x$

Square both sides: .$\displaystyle (2\sin x - 3)^2 \;=\;(3\cos x)^2$

. . . . . . . . . .$\displaystyle 4\sin^2\!x - 12\sin x + 9 \;=\;9\cos^2\!x$

. . . . . . . . . .$\displaystyle 4\sin^2\!x - 12\sin x + 9 \;=\;9(1-\sin^2\!x)$

. . . . . . . . . .$\displaystyle 4\sin^2\!x - 12\sin x + 9 \;=\;9 - 9\sin^2\!x$

. . . . . . . . . . . . $\displaystyle 13\sin^2\!x - 12\sin x \;=\;0$

. . . . . . . . . . . $\displaystyle \sin x(13\sin x - 12) \;=\;0$

And we have:

. . $\displaystyle \sin x \:=\:0 \quad\Rightarrow\quad x \:=\: 180^on$

. . $\displaystyle 13\sin x - 12 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\tfrac{12}{13} \quad\Rightarrow\quad x \:=\:\sin^{-1}\left(\tfrac{12}{13}\right)$
. . . . $\displaystyle x \:=\:\begin{Bmatrix}67.38^o + 360^on \\ 112.62^o + 360^on \end{Bmatrix}$

Since squaring an equation often introduces extraneous roots,
. . we must check our results.

We find that the solutions are:

. . $\displaystyle \begin{Bmatrix}x \;=\;180^o + 360^on \\ \\[-4mm] x \;=\;112.62^o + 360^on\end{Bmatrix}\;\text{ for some integer }n$

• Jun 7th 2011, 08:19 AM
starshine84
I think I understand... thanks!