# finding all roots of a trignometric equation

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pooja sudha 2020 年 12 月 30 日
コメント済み: Ameer Hamza 2020 年 12 月 31 日
Can we find all roots of a trignometric equation using matlab?
for e.g., tan(x)-x=0.

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### 回答 (2 件)

Ameer Hamza 2020 年 12 月 30 日
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));
end
sols = uniquetol(sols, 1e-2);
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Ameer Hamza 2020 年 12 月 31 日
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));
end

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Image Analyst 2020 年 12 月 30 日
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);
end
You get
xIntersection =
-1.66533453693773e-16

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