He pensado que este subiendo, pero no sé cómo solucionarlo. Aquí está mi intento: <span class="math-container">$$\sin x-\sqrt{3}\ \cos x=1$ $</span> <span class="math-container">$$(\sin x-\sqrt{3}\ \cos x)^2=1$ $</span> <span class="math-container">$$\sin^2x-2\sqrt{3}\sin x\cos x\ +3\cos^2x=1$ $</span> <span class="math-container">% $ $$1-2\sqrt{3}\sin x\cos x\ +2\cos^2x=1$</span> <span class="math-container">$ de $$2\cos^2x-2\sqrt{3}\sin x\cos x=0$</span> <span class="math-container">$$2\cos x(\cos x-\sqrt{3}\sin x)=0$ $</span> <span class="math-container">$2\cos x=0\Rightarrow x\in {\frac{\pi }2(2n-1):n\in\Bbb Z}$</span>
Pero cómo se resuelve <span class="math-container">$$\cos x-\sqrt{3}\sin x=0$ $</span>