Solving for Variables contained an interval

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Joe
Joe 2023-3-11
评论： Joe 2023-3-11
How do I solve the equation y = (sin(x) * (2* cos(x) - 1)) / (1 + 2 * cos(x)) only for x in the intervall [0 1], because if solved for all x there are infinte solutions. Thank you all for your help!

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采纳的回答

Paul
Paul 2023-3-11
Ran in:
Referring to the form of y given in this comment:
syms x real
y = sin(x)*(2*cos(x) - 1) / ((1 + 2*cos(x)) * (1 - cos(x)))
y = 
fplot(y,[0 1]) % don't know why the plot shows up below and not here?
Find the inverse function
fx = (finverse(y))
fx = 
Sub so that we get a function of the form where we input y and output x
syms yy real
fx(yy) = subs(fx,x,yy)
fx(yy) = 
Imaginary part is zero
imag(fx)
ans(yy) = 
0
So all we need is the real part
fx = real(simplify(fx,100))
fx(yy) = 
Check a value
fx(25)
ans = 
copyobj(gca,figure)
hold on
yline(25);
xline(double(fx(25)))
axis([0 0.1 0 50])

更多回答（2 个）

Askic V
Askic V 2023-3-11
编辑：Askic V 2023-3-11
Ran in:
fun = @(x) (sin(x) .* (2.* cos(x) - 1)) ./ (1 + 2 .* cos(x)); % function
x0 = [0.1 2]; % initial interval
x = fzero(fun,x0)
x = 1.0472
t = linspace(0,2);
y = fun(t);
plot(t,y)
grid on
The function has value zero at x = 0, and the next one is at 1.0472. Therefore in the interval between 0 and 1 there is only a solution at 0.
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Joe
Joe 2023-3-11
编辑：Joe 2023-3-11
yes now i noticed the 2 y values for one x too. Since this is for a physics projekt I have a second equation I could use for which there is only one y value for one x in the interval [0 1]. It goes like this: y = sin(x)*(2*cos(x) - 1) / ((1 + 2*cos(x)) (1 - cos(x))). Also it doesn't matter that y vlaues are very small. It would also be possible to set y to an fixed value like 4 and slove for that. Thanks again for your help this would take me sooo much longer without your help
Walter Roberson
Walter Roberson 2023-3-11
Ran in:
syms x y real
eqn = y == sin(x)*(2*cos(x) - 1) / ((1 + 2*cos(x)) * (1 - cos(x)))
eqn = 
sol = solve(eqn, x, 'returnconditions', true)
sol = struct with fields:
x: [3×1 sym] parameters: k conditions: [3×1 sym]
sx = simplify(sol.x, 'steps', 20)
sx = 
sol.conditions
ans = 
b1L = solve(sx(1) == 0, sol.conditions(1), 'returnconditions', true)
b1L = struct with fields:
k: [0×1 sym] y: [0×1 sym] parameters: [1×0 sym] conditions: [0×1 sym]
b1U = solve(sx(1) == 1, sol.conditions(1), 'returnconditions', true)
b1U = struct with fields:
k: (log((exp(3i)*1i + 1i)^2/(exp(3i) - 1)^2 + ((exp(3i)*1i + 1i)*2i)/(exp(3i) - 1) - 1)*1i - log((exp(3i)*1i + 1i)^2/(exp(3i) - 1)^2 + 1)*1i + 3)/(6*pi) y: (exp(3i)*1i + 1i)/(exp(3i) - 1) parameters: [1×0 sym] conditions: symtrue
sk = simplify(b1U.k, 'steps', 20)
sk = 
0
sy = simplify(b1U.y, 'steps', 20)
sy = 
vpa(sy)
ans = 
b2L = solve(sx(2) == 0, sol.conditions(2), 'returnconditions', true)
b2L = struct with fields:
k: [0×1 sym] y: [0×1 sym] parameters: [1×0 sym] conditions: [0×1 sym]
b2U = solve(sx(2) == 1, sol.conditions(2), 'returnconditions', true)
b2U = struct with fields:
k: [0×1 sym] y: [0×1 sym] parameters: [1×0 sym] conditions: [0×1 sym]
b3L = solve(sx(3) == 0, sol.conditions(3), 'returnconditions', true)
b3L = struct with fields:
k: [0×1 sym] y: [0×1 sym] parameters: [1×0 sym] conditions: [0×1 sym]
b3U = solve(sx(3) == 1, sol.conditions(3), 'returnconditions', true)
b3U = struct with fields:
k: [0×1 sym] y: [0×1 sym] parameters: [1×0 sym] conditions: [0×1 sym]
So out of the three analytic branches for the equation, only one of them can be probed for boundaries. The first of them has no y boundary at x == 0 because the equation goes to infinity. It does have a y boundary at x == 1 of roughly 0.0709

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John D'Errico
John D'Errico 2023-3-11
编辑：John D'Errico 2023-3-11
Ran in:
syms x
y = (sin(x) * (2* cos(x) - 1)) / (1 + 2 * cos(x));
xsol = solve(y == 0)
xsol = 
There are only three primary solutions.
xsol = solve(y == 0,'returnconditions',true)
xsol = struct with fields:
x: [3×1 sym] parameters: k conditions: [3×1 sym]
As you can see, now solve treturns a more complete result.
xsol.x
ans = 
xsol.conditions
ans = 
So, for integer k, the set of all solutions is one of those given in xsol.x, parameterized by the integer value of k.
That first positive solution is at pi/3, whhich falls just slightly outside of the interval [0,1]. So the only solution in that interval is 0 itself.
pi/3
ans = 1.0472
And that is exactly what @Askic V told you. Best to just use fzero.

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