Solving a sin &#x2061;<!-- ⁡ --> x + b cos &#x2061;<!-- ⁡ --> x = c Suppose

Solving $a\mathrm{sin}x+b\mathrm{cos}x=c$
Suppose $a\mathrm{sin}x+b\mathrm{cos}x=c$ . My teacher told me that with changing variable we can solve it by this : $\left(c+b\right){t}^{2}-2at+\left(c-b\right)=0$ where $t=\mathrm{tan}\frac{x}{2}$ . I thought it is true always but today found a problem . When we define $\mathrm{sin}x$ and $\mathrm{cos}x$ with $\mathrm{tan}\frac{x}{2}$ , we should consider $\mathrm{cos}\frac{x}{2}\ne 0$ and $x\ne 2k\pi +\pi$ but it can be the answer of the main equation (i.e. $a\mathrm{sin}x+b\mathrm{cos}x=c$ ) ! So , that formula is incomplete ?
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use that
$\sqrt{{a}^{2}+{b}^{2}}\left(\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}\mathrm{sin}\left(x\right)+\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}\mathrm{cos}\left(x\right)\right)=c$
and
$\mathrm{cos}\left(\varphi \right)=\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}$
$\mathrm{sin}\left(\varphi \right)=\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}$
therefore we get
$\mathrm{sin}\left(x+\varphi \right)=\frac{c}{\sqrt{{a}^{2}+{b}^{2}}}$
or you write
$2\phantom{\rule{thinmathspace}{0ex}}\frac{a\mathrm{tan}\left(x/2\right)}{1+{\left(\mathrm{tan}\left(x/2\right)\right)}^{2}}+\frac{b\left(1-{\left(\mathrm{tan}\left(x/2\right)\right)}^{2}\right)}{1+{\left(\mathrm{tan}\left(x/2\right)\right)}^{2}}=c$
with
$\mathrm{tan}\left(x/2\right)=t$