Question

помогите пожалуйста, ничего не понимаю​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{Sin(\pi -\alpha )Cos(\pi +\alpha )tg(-\alpha )}{Sin(\alpha -\frac{3\pi }{2} )Ctg(\frac{3\pi }{2} +\alpha ) Cos(\alpha +\frac{\pi }{2}) }=\frac{Sin\alpha \cdot(-Cos\alpha )\cdot(-tg\alpha )}{Cos\alpha \cdot(-tg\alpha )\cdot(-Sin\alpha )} =1\\\\\\2)\\\\2Cos^{2} 0,2x+\sqrt{3} Cos0,2x=0\\\\Cos0,2x\cdot(2Cos0,2x+\sqrt{3} )=0\\\\\\\left[\begin{array}{ccc}Cos0,2x=0\\2Cos0,2x+\sqrt{3}=0 \end{array}\right$

$\displaystyle\bf\\\left[\begin{array}{ccc}Cos0,2x=0\\Cos0,2x=-\dfrac{\sqrt{3} }{2} \end{array}\right\\\\\\1)\\\\Cos0,2x=0\\\\0,2x=\frac{\pi }{2} +\pi n,n\in Z\\\\\boxed{x=\frac{5\pi }{2} +5\pi n,n\in Z}\\\\2)\\\\Cos0,2x=-\frac{\sqrt{3} }{2} \\\\0,2x=\pm \ arcCos\Big(-\frac{\sqrt{3} }{2} \Big)+2\pi n,n\in Z\\\\0,2x=\pm \ \frac{5\pi }{6} +2\pi n,n\in Z\\\\\boxed{x=\pm \ \frac{25\pi }{6} +10\pi n,n\in Z}$