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I have an equation that needs to be solved numerically.
-125=-arctan(0.25*x)-0.2*x*180/pi
I know that the answer can be approximatedto 6, according to the exams "corrected answer"
I cant solve it with pen and paper so I wonder if there is anyway to solve this in MATLAB?

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Star Strider
Star Strider 2021-3-25

0 个投票

Here is how I would solve it:
fcn = @(x) 125-atand(0.25.*x)-0.2*x*180/pi;
est_x = fzero(fcn, 10)
producing:
est_x =
5.9959
See the documentation for fzero for details.

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