Mathematics - Trigonometric Equation Question with Solution | TestHub

Let $N(x)=2\cos x-1$ and $D(x)=2\sin x-\sqrt{3}$. We need to find $x\in [0,2\pi ]$ such that $\frac{N(x)}{D(x)}\ge 0$. Critical points are where $N(x)=0$ or $D(x)=0$.$N(x)=0⟹\cos x=1/2⟹x=\pi /3,5\pi /3$.$D(x)=0⟹\sin x=\sqrt{3}/2⟹x=\pi /3,2\pi /3$. Since $D(x)$ is in the denominator, $D(x)\ne 0$, so $x\ne \pi /3$ and $x\ne 2\pi /3$. Analyze the signs of $N(x)$ and $D(x)$ in $[0,2\pi ]$:$N(x)>0$ for $x\in [0,\pi /3)\cup (5\pi /3,2\pi ]$.$N(x)<0$ for $x\in (\pi /3,5\pi /3)$.$D(x)>0$ for $x\in (\pi /3,2\pi /3)$.$D(x)<0$ for $x\in [0,\pi /3)\cup (2\pi /3,2\pi ]$. Combining the signs for $\frac{N(x)}{D(x)}\ge 0$: 1. For $x\in [0,\pi /3)$: $N(x)>0,D(x)<0⟹N(x)/D(x)<0$. 2. For $x\in (\pi /3,2\pi /3)$: $N(x)<0,D(x)>0⟹N(x)/D(x)<0$. 3. For $x\in (2\pi /3,5\pi /3)$: $N(x)<0,D(x)<0⟹N(x)/D(x)>0$. 4. For $x\in (5\pi /3,2\pi ]$: $N(x)>0,D(x)<0⟹N(x)/D(x)<0$. We must include points where $N(x)=0$ but $D(x)\ne 0$. At $x=5\pi /3$, $N(x)=0$ and $D(5\pi /3)=2\sin (5\pi /3)-\sqrt{3}=2(-\sqrt{3}/2)-\sqrt{3}=-2\sqrt{3}\ne 0$. So $x=5\pi /3$ is a solution. Therefore, the solution set is $(2\pi /3,5\pi /3]$.