Mathematics - Trigonometric Equation Question with Solution | TestHub

The given inequality is $\cos x-\sin x\ge 1$. This can be rewritten using the amplitude-phase form as $\sqrt{1^2+(-1{)}^2}\cos (x+\frac{\pi }{4})\ge 1$, which simplifies to $\sqrt{2}\cos (x+\frac{\pi }{4})\ge 1$, or $\cos (x+\frac{\pi }{4})\ge \frac{1}{\sqrt{2}}$. Let $y=x+\frac{\pi }{4}$. Since $x\in [0,2\pi ]$, $y\in [\frac{\pi }{4},2\pi +\frac{\pi }{4}]=[\frac{\pi }{4},\frac{9\pi }{4}]$. In this interval, $\cos y\ge \frac{1}{\sqrt{2}}$ when $y=\frac{\pi }{4}$ or $y\in [\frac{7\pi }{4},\frac{9\pi }{4}]$. For $y=\frac{\pi }{4}$, we have $x+\frac{\pi }{4}=\frac{\pi }{4}⟹x=0$. For $y\in [\frac{7\pi }{4},\frac{9\pi }{4}]$, we have $x+\frac{\pi }{4}\in [\frac{7\pi }{4},\frac{9\pi }{4}]$, which implies $x\in [\frac{7\pi }{4}-\frac{\pi }{4},\frac{9\pi }{4}-\frac{\pi }{4}]=[\frac{3\pi }{2},2\pi ]$. So the solutions are $x=0$ and $x\in [\frac{3\pi }{2},2\pi ]$. We need to find the number of integral values of $x$. For $x=0$, it is an integer. For $x\in [\frac{3\pi }{2},2\pi ]$, we approximate the values: $\frac{3\pi }{2}\approx 1.5\times 3.14159=4.712$ and $2\pi \approx 2\times 3.14159=6.283$. The integers in $[4.712,6.283]$ are $5$ and $6$. Thus, the integral values of $x$ are $0,5,6$. The total number of integral values is 3.