# Thread: roots of the equation

1. ## roots of the equation

Find the roots of the equations:
1. $sin(x)=x$
2. $cos(x)=x$
3. $tan(x)=x$

2. 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.

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

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.

4. 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 $\pi$ though?

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

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.

$\sin{x} = x$
$\cos{x} = 1-\frac{x^2}{2}$
$\tan{x} = x$

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

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

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