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.



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