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 $latex t$:
$$ -6 t ^ 2 +8 t= 0$$
$$ 3t ^ 2-4t= 0$$
Which is addressed by factoring:
$$ 3t ^ 2-4t= 0Rightarrow t( 3t-4)= 0$$
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.