Trigonometric equation without graphs

Question

$$\cos a-\sin a\ge0,\ a\in[0,2\pi]$$ $\cot a\ge1$. After this I am unable to proceed, is there a method to calculate the answer without drawing a graph?

Answer 1

$$\frac{\sqrt 2}{2}\cos a-\frac{\sqrt 2}{2}\sin a\ge 0$$ $$\cos a\cos\frac{\pi}{4}-\sin a\sin\frac{\pi}{4}\ge 0$$ $$\cos(a+\frac{\pi}{4})\ge 0$$

Call $b=a+\frac{\pi}{4}$, then $b\in[\frac{\pi}{4}, 2\pi+\frac{\pi}{4}]$ and $\cos b\ge 0$ for $b\in[\frac{\pi}{4}, \frac{\pi}{2}]\cup[\frac{3\pi}{2},2\pi+\frac{\pi}{4}]$. Then, $a\in[0,\frac{\pi}{4}]\cup[\frac{5\pi}{4}, 2\pi]$.

Answer 2

$$\cos a-\sin a=\sqrt2\left(\frac1{\sqrt2}\cos a-\frac1{\sqrt2}\sin a\right)=$$ $$==\sqrt2\left(\cos\frac{\pi}4\cos a-\sin\frac{\pi}4\sin a\right)=\sqrt2\cos \left(a+\frac{\pi}4\right)$$

Answer 3

$$a\in(0,\pi)\implies\sin a>0$$ allows you to write $$\cot a\ge1$$ or $$a\ge\text{arccot }1=\frac\pi4.$$

Similarly,

$$a\in(\pi,2\pi)\implies\sin a<0$$ allows you to write $$\cot a\le1$$ or $$a\le\pi+\text{arccot }1=\frac{3\pi}4.$$

Hence, adding the values at endpoints of the intervals,

$$\left[\frac\pi4,\frac{3\pi}4\right].$$