# roots of the equation

• Feb 27th 2009, 10:58 PM
u2_wa
roots of the equation
Find the roots of the equations:
1. $\displaystyle sin(x)=x$
2. $\displaystyle cos(x)=x$
3. $\displaystyle tan(x)=x$
• Feb 27th 2009, 11:10 PM
sbcd90
draw the graphs in each case to find solutions.

suppose,y=sinx & y=x gives the solution of first equation.
though sin(x) becomes x when x is a very small proper fraction.
• Feb 28th 2009, 01:55 AM
mr fantastic
Quote:

Originally Posted by u2_wa
Find the roots of the equations:
1. $\displaystyle sin(x)=x$
2. $\displaystyle cos(x)=x$
3. $\displaystyle tan(x)=x$

These are transcendental equations - which means that exact solutions cannot be found except for special cases eg. For 1. and 3. x = 0 is one of the solutions. To get the other solutions you will need to use technology to get decimal approximations of them.

Quote:

Originally Posted by sbcd90
draw the graphs in each case to find solutions.

suppose,y=sinx & y=x gives the solution of first equation.

though sin(x) becomes x when x is a very small proper fraction.

This is only good if either:

1. You don't want very good accuracy in your solutions, or

2. You use technology to draw them and then get that same technology to find the x-coordinates of the intersection points.
• Feb 28th 2009, 04:55 AM
e^(i*pi)
Quote:

Originally Posted by mr fantastic
These are transcendental equations - which means that exact solutions cannot be found except for special cases eg. For 1. and 3. x = 0 is one of the solutions. To get the other solutions you will need to use technology to get decimal approximations of them.

You could also leave them in terms of $\displaystyle \pi$ though?

For example $\displaystyle \sin{x} = 0$ would have solutions of $\displaystyle 0 + {k}\pi$ where k is an integer

Quote:

Originally Posted by sbcd90
though sin(x) becomes x when x is a very small proper fraction.

Note this only words if x is in radians, the small angle approximation (what sbcd90 describes) is a truncation of the Taylor series for sin, cos and tan respectively. It works because as x is much less than one it's powers will decrease rapidly.

$\displaystyle \sin{x} = x$
$\displaystyle \cos{x} = 1-\frac{x^2}{2}$
$\displaystyle \tan{x} = x$
• Feb 28th 2009, 12:15 PM
mr fantastic
Quote:

Originally Posted by e^(i*pi)
You could also leave them in terms of $\displaystyle \pi$ though?

For example $\displaystyle \sin{x} = 0$ would have solutions of $\displaystyle 0 + {k}\pi$ where k is an integer

[snip]

$\displaystyle x = k \pi$ is not a solution to $\displaystyle \sin x = x$. The only solution of this form is $\displaystyle x = 0$.