matlab 中acos,Matlab：返回反余弦（acos）的完整解(Matlab: return complete solution of inverse cosine (acos))...

Here's a numerical solution; like Benoit_11 I don't see the point of doing it symbolically in this context.

There are two solutions within the interval [-pi, pi], the larger one being returned by acos:

% solution within [0, pi]

theta1 = acos(x);

% solution within [-pi, 0]

theta2 = -acos(x);

These solutions repeat at steps of 2 pi. The number of possible steps downwards and upwards can be determined by the integer part of the distance between the basic solution and the respective limit (lower and upper), in units of 2 pi. For theta1:

% repetitions in 2 pi intervals within limits

sd = floor((theta1 - lower) / (2 * pi));

su = floor((upper - theta1) / (2 * pi));

theta1 = (-sd : su) * 2 * pi + theta1

And the same for theta2:

% repetitions in 2 pi intervals within limits

sd = floor((theta2 - lower) / (2 * pi));

su = floor((upper - theta2) / (2 * pi));

theta2 = (-sd : su) * 2 * pi + theta2

If you'd like one combined list of solutions, excluding possible duplicates:

theta = unique([theta1, theta2])

and in degrees:

theta = theta / pi * 180;

Example:

x = 0.5;

lower = -10;

upper = 30;

gives

theta =

-7.3304 -5.2360 -1.0472 1.0472 5.2360 7.3304 11.5192 13.6136 17.8024 19.8968 24.0855 26.1799