Question

корень из sinx×cosx ×(1/tg^2x+1)=0

Answer 1

$\sqrt{sinx\cdot cosx}\cdot \frac{1}{tg^2x+1}=0\; ,\\\\ODZ:\; \; \left \{ {{sinx\cdot cosx \geq 0} \atop {tg^2x+1\ne0,\; x\ne \frac{\pi}{2}+\pi n}} \right. \; \left \{ {{\frac{1}{2}sin2x \geq 0} \atop {tg^2x\ne -1}} \right. \; \left \{ {{2\pi n \leq 2x \leq \pi +2\pi n} \atop {x\ne \frac{\pi}{2}+\pi n}} \right. \; \to \\\\\pi n \leq x \ \textless \ \frac{\pi}{2}+\pi n\; ,\; n\in Z\\\\1)\; \; sinx\cdot cosx=\frac{1}{2}sin2x=0\\\\2x=\pi n\; ,\; \; x=\frac{\pi n}{2}\; ,\; \; n\in Z$

$2)\; \; \left \{ {{\pi n \leq x\ \textless \ \frac{\pi}{2}+\pi n} \atop {x=\frac{\pi n}{2}}} \right. \; \; \to \; \; x=\pi n,\; n\in Z$