The number of solutions of the equation | x| = x+1 x in the interval [0,2] is ___________ …

Question

The number of solutions of the equation $|\cot x| = \cot x+\dfrac1{\sin x}$ in the interval $[0,2\pi]$ is \_\_\_\_\_\_\_\_\_\_\_

Answer: 1

Accepted Answer

If $\cot x>0$: $\cot x = \cot x+\dfrac1{\sin x} \implies \dfrac1{\sin x}=0$, impossible.

If $\cot x<0$: $-\cot x = \cot x+\dfrac1{\sin x} \implies 2\cot x+\dfrac1{\sin x}=0 \implies 2\cos x=-1 \implies x=\dfrac{2\pi}3 \text{ or } \dfrac{4\pi}3$

At $x=\dfrac{4\pi}3$ (third quadrant), $\cot x>0$, contradicting the case assumption, so it's rejected. Only $x=\dfrac{2\pi}3$ is valid.

$$\text{Number of solutions} = 1$$

Solutions · 1