To address this formula, which consists of an independent term apart from \$latex 0\$, the list below modification of variable is utilized:

\$\$ t= tan left( dfrac {x} {2} right)\$\$

Thus, sine as well as cosine are shared as adheres to:

\$\$ cos x= dfrac {1-t ^ 2} {1+ t ^ 2} \$\$

\$\$ wrong x= dfrac {2t} {1+ t ^ 2} \$\$

This modification of variable is referred to as the global replacement or Weierstrass replacement. After that, the initial formula appears like this:

\$\$ 4sin x +3 cos x = 3Rightarrow 4left( frac {2t} {1+ t ^ 2} right) +3 left( frac {1-t ^ 2} {1+ t ^ 2} right)= 3\$\$

\$\$ frac {8t +3 -3 t ^ 2} {1+ t ^ 2} =3Rightarrow 8t +3 -3 t ^ 2= 3 +3 t ^ 2\$\$

Regrouping the terms, we acquire a square formula in \$latex t\$:

\$\$ -6 t ^ 2 +8 t= 0\$\$

\$\$ 3t ^ 2-4t= 0\$\$

\$\$ 3t ^ 2-4t= 0Rightarrow t( 3t-4)= 0\$\$

The options are:

\$ latex t_1 =0\$

\$ latex t_2= dfrac {4} {3} \$

This cause 2 basic formulas, by replacing back:

Formula 1

\$\$ tan left( dfrac {x} {2} right)= 0\$\$

Therefore:

\$ latexdfrac {x} {2} =arctan 0\$

\$ latex dfrac {x} {2} =0, pmpi, pm 2pi, pm 3pi, …\$

\$ latex x= 0, pm 2pi, pm 4pi, pm 6pi, …\$

\$ latex x= 2kpi= 2kcdot 360º\$

Where \$latex k\$ is an integer.

Formula 2

\$\$ tan left( dfrac {x} {2} right)= dfrac {4} {3} \$\$

We reason that:

\$ latex dfrac {x} {2} =arctan left( dfrac {4} {3} right)= 53.1 º\$

Therefore, the major option, which remains in the interval\$ latex 0leq xleq 360º \$, is:

\$ latex x= 106.2 º\$

Furthermore, the angles \$latex -253.8 º, 466.2 º, …\$ are likewise options of the suggested formula, considering that the tangent feature is routine, consequently, as a whole kind, the option is provided by:

\$ latex x = kcdot 360º+ 106.2 º\$

Where \$latex k = 0, pm 1, pm 2, pm 3 …\$, that is, \$latex k\$ is an integer.