Finding epsilon satisfy a trigonometric function.

I need to find the value of epsilon (e) that satisfy the function cot(e)^2*cot(2*e)^2 == sin(2*e)^2 where pi/4<e<pi/2

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sawsan
sawsan am 31 Dez. 2016
Bearbeitet: Walter Roberson am 3 Jan. 2017
Thanks a lot. I found the solution. I share it here if someone want to use it.
syms x
f = cot(x).^2*cot(2*x).^2 - sin(2*x).^2;
vpasolve(f,[pi/3 pi/2])
ans =
1.326722925756775867399902790294

3 Kommentare

There are 4 solutions on the interval (-pi/2,pi/2).
yes, but on the interval [pi/3, pi/2] there is only one solution. Thanks for replying.
Your original Question asked for all solutions on the interval -pi/2 to pi/2.

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Weitere Antworten (2)

KSSV
KSSV am 27 Dez. 2016
clc; clear all ;
syms e
eqn = cot(e)^2*cot(2*e)^2 == sin(2*e)^2 ;
[solx, params, conds] = solve(eqn, e, 'ReturnConditions', true) ;
Star Strider
Star Strider am 27 Dez. 2016
Another approach:
trig_e = @(e) cot(e).^2*cot(2*e).^2 - sin(2*e).^2;
v = linspace(-pi/2, pi/2, 4);
for k1 = 1:length(v)
e(k1) = fzero(trig_e, v(k1));
end
e =
-1.3267e+000 -541.3248e-003 541.3248e-003 1.3267e+000

Gefragt:

am 27 Dez. 2016

Kommentiert:

am 3 Jan. 2017

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