finding all roots of a trignometric equation

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Can we find all roots of a trignometric equation using matlab?
for e.g., tan(x)-x=0.

Answers (2)

Ameer Hamza
Ameer Hamza on 30 Dec 2020
range of tan(x) is (-inf inf), so this equation has an infinite number of solutions. Also, the solutions to this equation cannot be represented analytically. There is no general way to find multiple solutions to such equations. One solution is to start the numerical solver with several starting points and choose unique values. For example
eq = @(x) tan(x)-x;
x_range = 0:0.1:20;
sols = ones(size(x_range));
for i = 1:numel(sols)
sols(i) = fsolve(eq, x_range(i));
sols = uniquetol(sols, 1e-2);
Walter Roberson
Walter Roberson on 30 Dec 2020
The solutions to tan(x)-x tend towards being close to 2*pi apart, so it is possible to generate reasonable starting points for as many points as desired until you start losing too much floating point precision.
However, you will never have enough memory to return them "all".
Ameer Hamza
Ameer Hamza on 31 Dec 2020
Yes, thats a good observation to give the initial points. The solutions are pi apart from each. Following will also give same result as one in the answer.
eq = @(x) tan(x)-x;
x_range = 0.1:pi:20;
sols = ones(size(x_range));
for i = 1:numel(sols)
sols(i) = fsolve(eq, x_range(i));

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Image Analyst
Image Analyst on 30 Dec 2020
Well here's one way. Use fminbnd() to find out where the equation is closest to zero:
x = linspace(-1, 1, 1000);
y = tan(x);
plot(x, x, 'b-', 'LineWidth', 2);
hold on;
plot(x, y, 'r-', 'LineWidth', 2);
grid on;
axis equal
% Use fminbnd() to find where d = 0, which is where tan(x) = x.
xIntersection = fminbnd(@myFunc, -1.5,1.5)
% Put a line there
xline(xIntersection, 'Color', 'g', 'LineWidth', 3);
caption = sprintf('xIntersection = %f', xIntersection);
title(caption, 'FontSize', 20)
function d = myFunc(x)
d = abs(tan(x) - x);
You get
xIntersection =


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