error using solve function

hi,

I use the function solve to determine a variable as follows:

x = solve (cos (x) * sin (y) * cos (z) + sin (Z) = 1 ',' x ')

then I evaluated x in terms of y and z, when I replace x by the value finding the equation is not = 1!!

how can i resolve it?

Answer 1

[Walter Roberson]

Are x and y and z symbolic at the time you call solve? Or are y and z numeric?

If they are symbolic and you later substitute in numeric y and z to the solve() response get a numeric x, and then substitute the numeric x, y, and z into the cos (x) * sin (y) * cos (z) + sin (Z) formula, then you could get something other than 1 due to round-off error.

Also, if your y and z are square matrices of numeric values, then cos (x) * sin (y) * cos (z) would succeed but what not be what you want: you would need to evaluate

cos(x) .* sin(y) .* cos(z)

Notice the ".*" instead of "*" if you are working with vectors or matrices.

[studentU]

thank's for your repply,

the x and y are the symbolic when i call solve,

you're talking about ".*" in the verification? or at the eval function?

[Walter Roberson]

When you do the numeric verification, if you are testing with vectors of values, then you will need to use .* vectorization when you write the formulas. Or you could use matlabFunction() on the symbolic expression as that will automatically vectorize.

If you are doing symbolic verification instead of numeric, and you do not get 1, then you might need to simplify() the result.

[studentU]

I used simplify () but the result is always diferrent 1,

However, how to limit the interval of solution; for

 x = solve ('cos (x) * sin (y) * cos (z) + sin (Z) = 1 ',' x ') 
i want x beteween 0° and 90° for exmple,
thank's lot,

[Torsten]

I'd be interested what you get for x after applying "solve".

Best wishes

Torsten.