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BOBIN SAJI GEORGE on 18 Feb 2016
Answered: Walter Roberson on 18 Feb 2016
5.7938+8.961y-8.961ycos(0.5t)-1.88sin(0.5t)=0
1.867y-1.8817ycos(0.5t)+8.957ysin(0.5t)+1.4142=0

Moe_2015 on 18 Feb 2016
I would recommend that you read MathWorks documentation as it is very helpful. Here is a link where they show you how to solve a system of nonlinear equations with fsolve:

Walter Roberson on 18 Feb 2016
There is not just one solution, there are 4 (or more solutions).
There are 4 solutions if you assign precise values to each of the coefficients given as floating point values, such as if you change each of them to a rational value or if the final coefficient in the second equation should be sqrt(2) . If you can do that conversion then the solution involves a quartic (polynomial of degree 4), for which there are closed-form solutions... solutions that are pretty long to write down and almost incomprehensible when they are written down. Two of the solutions are real-valued and two are imaginary-valued.
However, considering that the floating point values are written down with anywhere from 1 to 5 significant digits, especially with the 1.4142 looking so suggestive of sqrt(2), any mathematician or scientist or engineer would have to assume that the values given are not exact values, and that instead need to be understood as rounded values with the true values being anything that rounds to the same value. That, for example, 5.7938 is not truly 57938/10000 but instead is 57938/10000 +/- 5/100000. When you make this substitution of ranges instead of exact values, then you get ranges of solutions instead of particular solutions.
I used a symbolic package to substitute rational intervals +/- 10^(-5) wide for each of the floating point coefficients, except that I used 0.5 as 1/2 exactly instead of as an interval. When those intervals are substituted and you solve(), the leading coefficient for the quartic is a range that crosses 0. Experimenting, the t values turn out to be pretty stable, but I find y values between -2.8046331179521356489*10^6 and 7.2637133118388977883*10^5 . The calculations appear to be especially sensitive to the error term used for the 8.961y (but other entries also have significant influence.)
The situation is effectively a case of Garbage In, Garbage Out (GIGO). Looking for meaningful solutions of non-linear equations with floating-point coefficients is often a waste of time as often the coefficients provided turn out to be insufficiently precise to warrant an answer.