The equation I'm trying to solve is this:
$$B\times \tan \phi - \frac{1}{2}g \left ( \frac{B}{v_{0}\cos\phi} \right )^{2}=A$$
$\textrm{A and B are constants the unknown is angle phi}$
I don't know if Casio CFX 9860 has the capability to solve such equation. I'm aware that Maple can do the job, but it is not what I'm looking for.
Can somebody help me if there is a program already written for solving this kind of equations.
Let's say if:
$\textrm{A= 16, B= 20}$
$v_{0}=20$
I tried to simplify the equation in such a way that it can be put in terms of one trigonometric expression but the cosine which is in the denominator does seem to be difficult to reduce.
I have made two approaches to solve this problem, one approach leads to a quartic equation and the other quadratic equation. $$B\times \tan \phi - \frac{1}{2}g \left ( \frac{B}{v_{0}\cos\phi} \right )^{2}=A\tag{1}$$
Quadratic Equation: $B^2gx^2-2v_0^2Bx+(B^2g+A2v^2)=0$
Where $x=\tan\phi$
Simplifying $(1)$ gives me: $$B \tan \phi - \frac{gB^2}{2v_0^2\cos^2\phi}=A\Rightarrow B\tan\phi-\frac{gB^2\sec^2\phi}{2v_0^2}$$ Since $\tan^2\phi+1=\sec^2\phi$, then $$B\tan\phi\cdot2v_0^2-B^2g(\tan\phi+1)=A\cdot2v_0^2$$ $$2v_o^2B\tan\phi-B^2g\tan^2\phi-B^2g-2v_0^2A=0$$ Letting $x=\tan\phi$, we now have a quadratic equation: $$B^2gx^2-2v_0^2Bx+(B^2g+2v_0^2A)=0$$ $$x\Rightarrow\frac{2v_0^2B\pm\sqrt{(2v_0^2B)^2-4(B^2g)(B^2g+2v_0^2A)}}{2B^2g}$$ And thus: $$\phi\Rightarrow\tan^{-1}\frac{2v_0^2B\pm\sqrt{(2v_0^2B)^2-4(B^2g)(B^2g+2v_0^2A)}}{2B^2g}$$
Quartic Equation: $(4v_0^4B)x^4-(4AgBv_0^2-4v_0^4B)x^2+(4A^2v_0^4)x-g^2B^2=0$
where $x=\cos \phi$ Simplifying $(1)$ gives me: $$B\frac{\sin\phi}{\cos\phi}-\frac{gB^2}{2v_0^2\cos^2\phi}=A$$ $$\frac{B\sin\phi\cdot2v_0^2\cos\phi-gB^2}{2v_0^2\cos^2\phi}=A$$ $$B\sin\phi\cdot2v_0^2\cos\phi=gB^2+A\cdot2v_0^2\cos^2\phi\tag{2}$$ Now to eliminate $\sin\phi$, we need to square $(2)$ and thus: $$4v_0^4\sin^2\phi\cos^2\phi=g^2B^2+4A^2v_0^4\cos^4\phi+4AgBv_0^2\cos^\phi$$ By the Pythagorean identity $\cos^2\phi+\sin^2\phi=1$, we can rewrite the equation above as: $$4v_0^4B(1-\cos^2\phi)\cos^2\phi=g^2B^2+4A^2v_0^4\cos^4\phi+4AgBv_0^2\cos^\phi$$ $$4v_0^4B\cos^2\phi-4v_0^4B\cos^4\phi=g^2B^2+4A^2v_0^4\cos^4\phi+4AgBv_0^2\cos^\phi$$ And thus rewriting the equation above, letting $x=\cos\phi$, we have: $$4v_0^4Bx^4-(4AgBv_0^2-4v_0^4B)x^2+4A^2v_0^4x-g^2B^2=0$$ Maybe then you can use Ferrari's method to solve quartic formulas.