How to I find x in sin(x) = 0.65

Im new to matlab and am fine using a solve equation but when trying to find the answer to something with infinite solutions nothing occurs, I dont know how to add the domain the this equation and would like some assistance please

Answer 1

[John D'Errico]

Simple enough. I'll use solve, since you mentioned that. What happens if you just call solve?

syms x
solve(sin(x) == 0.5)

ans = 

So there are two fundamentally different solutions. Of course, asin can return only one of them, and since the range of the function asin is [-pi/2,pi/2], then asin would return the solution at pi/6, or 30 degrees.

If you wanted to find all solutions, then add a qualifier to solve.

sol = solve(sin(x) == 0.5,'returnConditions',true)

sol = struct with fields:
             x: [2×1 sym]
    parameters: k
    conditions: [2×1 sym]

sol.x

ans = 

sol.conditions

ans = 

So for integer values of k, the general solutions are the primary solutions, plus any integer multiple of 2*pi.

If you wanted to solve this in degrees, that becomes

sol = solve(sind(x) == 0.5,'returnConditions',true)

sol = struct with fields:
             x: [2×1 sym]
    parameters: k
    conditions: [2×1 sym]

sol.x

ans = 

There are primary solutions at 30 and 150 degrees, plus any integer multiple of 360 degrees.

Answer 2

[Arif Hoq]

a=sind(30)

a = 0.5000

x=asind(0.5) %Inverse sine in degrees

x = 30.0000

Answer 3

[Sam Chak]

Hi @Eric Seinen

Not trying to confuse you, but here is the 3rd approach, the Numerical Approach.

fun = @(x) sin(x) - 0.65;   % function, fun(x) = 0
x0 = 0;                     % initial point
x = fzero(fun, x0)          % find the root when fun(x) = 0

x = 0.7076

Since the sine function is periodic, there are infinitely many solutions. However, there are two solutions in the principal domain . So, if you choose another initial point that is apart, then you will get another solution.

fun = @(x) sin(x) - 0.65;   % function, fun(x) = 0
x0 = pi;                    % initial point
x = fzero(fun, x0)          % % find the root when fun(x) = 0

x = 2.4340

@Arif Hoq's method is called the Inverse Function approach. The inverse of the sine function is called "arcsine". So, asin(y) can be used. It will return the value of x in radian.

However, sometimes certain equations are transcendental equations, where the equation cannot be solved for x algebraically. Because of this, the numerical approach is employed, which requires an iterative, a priori (initial-point-guessing) procedure. For example, the Kepler's equation:

.

@John D'Errico's method is called the Symbolical Approach, in which the solution is returned in the form of symbols.