Solving for Variables contained an interval

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!

Answer 1

[Paul]

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])

[Joe]

Thanks you gave me exactly what I need

Answer 2

[Askic V]

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.

[Joe]

First, thank you for your time. But, as you can clearly see there are y values for x = 0.5 and so on. I am looking for an equation that yields, for a given y value, a x value between 0 and 1. Between 0 and 1, because the function has infinite solutions for an non restriced interval

[Torsten]

But the range of y-values the function takes in [0 1] is very limited as you can see from its graph ...

And for most of these y-values, there are two x-values x1, x2 with f(x1) = f(x2) = y.

[Joe]

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]

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

Answer 3

[John D'Errico]

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.