Question

$$md\times1f\ \cos(3x-30^{\circ})=s\ in(7x+50^{\circ})$$

Answer

m=sin(7*x+50*deg)/(d*f*cos(3*x-30*deg))

Solution


Simplify  \(md\times 1\times f\cos{(3x-30deg)}\)  to  \(mdf\cos{(3x-30deg)}\).
\[mdf\cos{(3x-30deg)}=\sin{(7x+50deg)}\]
Divide both sides by \(d\).
\[mf\cos{(3x-30deg)}=\frac{\sin{(7x+50deg)}}{d}\]
Divide both sides by \(f\).
\[m\cos{(3x-30deg)}=\frac{\frac{\sin{(7x+50deg)}}{d}}{f}\]
Simplify  \(\frac{\frac{\sin{(7x+50deg)}}{d}}{f}\)  to  \(\frac{\sin{(7x+50deg)}}{df}\).
\[m\cos{(3x-30deg)}=\frac{\sin{(7x+50deg)}}{df}\]
Divide both sides by \(\cos{(3x-30deg)}\).
\[m=\frac{\frac{\sin{(7x+50deg)}}{df}}{\cos{(3x-30deg)}}\]
Simplify  \(\frac{\frac{\sin{(7x+50deg)}}{df}}{\cos{(3x-30deg)}}\)  to  \(\frac{\sin{(7x+50deg)}}{df\cos{(3x-30deg)}}\).
\[m=\frac{\sin{(7x+50deg)}}{df\cos{(3x-30deg)}}\]