+893 I don't think that you're going to get very far with this, unless there happens to be some nice relationship between x and y.
The way to deal with it is to take the cos(a) into the sq root as sec^2(a) and to replace it with 1+tan^2(a). That creates a cubic in tan(a) which will remain, even when the square root is removed.
+118677 $$\\V=\frac{(2.21x)}{cos(a)}*\sqrt{-y+tan(a)*x}\\
V^2=\frac{(2.21x)^2}{cos^2(a)}*(-y+tan(a)*x)\\
V^2cos^2a=(2.21x)^2*(-y+tan(a)*x)\\
V^2cos^2a=-(2.21x)^2y+(2.21x)^2tan(a)*x\\
V^2cos^2a+(2.21x)^2y=(2.21x)^2tan(a)*x\\
V^2cos^2a+(2.21x)^2y=(2.21x)^2\sqrt{\frac{1}{cos^2a}-1}*x\\
V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1}{cos^2a}-1)}*x^2\\
V^4cos^4a+2*V^2cos^2a*(2.21x)^2y+(2.21x)^4y^2=(2.21x)^4*(\frac{1-cos^2a}{cos^2a})*x^2\\
V^4cos^6a+2*V^2cos^4a*(2.21x)^2y+(2.21x)^4y^2cos^2a=(2.21x)^4*({1-cos^2a})*x^2\\$$
I am not getting anywhere with this. Sorry.
+893 I don't think that you're going to get very far with this, unless there happens to be some nice relationship between x and y.
The way to deal with it is to take the cos(a) into the sq root as sec^2(a) and to replace it with 1+tan^2(a). That creates a cubic in tan(a) which will remain, even when the square root is removed.
