Filter löschen
Filter löschen

not getting proper solution

2 Ansichten (letzte 30 Tage)
Kanav Vij
Kanav Vij am 30 Sep. 2015
Beantwortet: Walter Roberson am 30 Sep. 2015
i'm trying to solve this question but the solution is in different form, the solution should come as integer.
>> a=30;
>> m=30*pi/180;
>> syms x
>> eq= tan((2*x)-m)==(100-30*sin(x))/(100-30*cos(x));
>> solve(eq,x)
Warning: The solutions are parameterized by the symbols: k, z1. To include parameters and conditions in the solution,
specify the 'ReturnConditions' option.
> In solve>warnIfParams (line 510)
In solve (line 360)
Warning: The solutions are valid under the following conditions: 3*exp(- log(z1) - pi*k*2i) + 3*exp(log(z1) + pi*k*2i)
~= 20 & 2*3^(1/2)*(exp(- log(z1) - pi*k*2i)/2 + exp(log(z1) + pi*k*2i)/2)^2 + 2*(exp(- log(z1) - pi*k*2i)/2 +
exp(log(z1) + pi*k*2i)/2)*((exp(- log(z1) - pi*k*2i)*1i)/2 - (exp(log(z1) + pi*k*2i)*1i)/2) ~= 3^(1/2) & in(k,
'integer') & (z1 == RootOf(z^4 - z^3*(3/20 + 3i/20) - (z*(3^(1/2)*(3 - 3i) - (3 + 3i)))/40 + 3^(1/2)/2 - 1i/2, z)[1] |
z1 == RootOf(z^4 - z^3*(3/20 + 3i/20) - (z*(3^(1/2)*(3 - 3i) - (3 + 3i)))/40 + 3^(1/2)/2 - 1i/2, z)[2] | z1 ==
RootOf(z^4 - z^3*(3/20 + 3i/20) - (z*(3^(1/2)*(3 - 3i) - (3 + 3i)))/40 + 3^(1/2)/2 - 1i/2, z)[3] | z1 == RootOf(z^4 -
z^3*(3/20 + 3i/20) - (z*(3^(1/2)*(3 - 3i) - (3 + 3i)))/40 + 3^(1/2)/2 - 1i/2, z)[4]). To include parameters and
conditions in the solution, specify the 'ReturnConditions' option.
> In solve>warnIfParams (line 517)
In solve (line 360)
ans =
2*pi*k - log(z1)*1i

Antworten (1)

Walter Roberson
Walter Roberson am 30 Sep. 2015
plot tan((2*x)-m) - (100-30*sin(x))./(100-30*cos(x)) over a range of values. Do you see any crossings at integral x? If you do, then substitute that x into the equations and see if it is a true solution or only a near solution.
That equation has four real-valued non-rational solutions for every period of 2 Pi.

Kategorien

Mehr zu Linear Algebra finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by