[tex]2\sin\alpha-\cos\alpha=2[/tex]
Consider the substitution [tex]\tan\dfrac\alpha2=\beta[/tex]. Then by the double angle identities we get
[tex]\sin\alpha=2\sin\dfrac\alpha2\cos\dfrac\alpha2[/tex]
[tex]\cos\alpha=\cos^2\dfrac\alpha2-\sin^2\dfrac\alpha2[/tex]
We also have
[tex]\tan\dfrac\alpha2=\beta\implies\begin{cases}\sin\dfrac\alpha2=\dfrac\beta{\sqrt{1+\beta^2}}\\\\\cos\dfrac\alpha2=\dfrac1{\sqrt{1+\beta^2}}\end{cases}[/tex]
so that
[tex]\sin\alpha=\dfrac{2\beta^2}{1+\beta^2}[/tex]
[tex]\cos\alpha=\dfrac{1-\beta^2}{1+\beta^2}[/tex]
and the original equation has been transformed to
[tex]\dfrac{4\beta^2-(1-\beta^2)}{1+\beta^2}=2[/tex]
Solve for [tex]\beta[/tex]:
[tex]5\beta^2-1=2+2\beta^2[/tex]
[tex]3\beta^2=3[/tex]
[tex]\beta^2=1[/tex]
[tex]\beta=\pm1[/tex]
Solving for [tex]\alpha[/tex] gives
[tex]\tan\dfrac\alpha2=-1\implies\dfrac\alpha2=-\dfrac\pi4+n\pi\implies\alpha=-\dfrac\pi2+2n\pi[/tex]
[tex]\tan\dfrac\alpha2=1\implies\dfrac\alpha2=\dfrac\pi4+n\pi\implies\alpha=\dfrac\pi2+2n\pi[/tex]
where [tex]n[/tex] is any integer. Both [tex]\sin[/tex] and [tex]\cos[/tex] are [tex]2\pi[/tex]-periodic, which is to say
[tex]\cos(x+2n\pi)=\cos x[/tex]
[tex]\sin(x+2n\pi)=\sin x[/tex]
so that
[tex]\sin\alpha=\sin\left(\pm\dfrac\pi2+2n\pi\right)=\sin\left(\pm\dfrac\pi2\right)=\pm1[/tex]
[tex]\cos\alpha=\cos\left(\pm\dfrac\pi2+2n\pi\right)=\cos\left(\pm\dfrac\pi2\right)=0[/tex]
and we find that
[tex]\sin\alpha+2\cos\alpha=\pm1[/tex]
