Question

Помогите пожалуйста решить, прошууууу (срочно) дам 60 баллов, хоть и мало (это все что есть)

Answer 1

$1)Sin\frac{x}{2}Cos\frac{x}{2}=\frac{1}{2}*2Sin\frac{x}{2}Cos\frac{x}{2}=\frac{1}{2}Sinx=0,5Sinx\\\\2)Cos^{4}x+Sin^{2}xCos^{2}x=Cos^{2}x(Cos^{2}x+Sin^{2}x)=Cos^{2}x*1=Cos^{2}x\\\\3)5Cos(\frac{\pi }{2}-\alpha)+Sin(\pi+\alpha)=5Sin\alpha-Sin\alpha=4Sin\alpha=4*(-0,4)=-1,6$

$4)Cos\alpha=-\frac{12}{13}\\\\1+tg^{2}\alpha=\frac{1}{Cos^{2}\alpha}\\\\tg^{2}\alpha=\frac{1}{Cos^{2}\alpha}-1=\frac{1}{(-\frac{12}{13})^{2}}-1=\frac{169}{144}-1=\frac{25}{144}\\\\tg\alpha=-\sqrt{\frac{25}{144}}=-\frac{5}{12}\\\\Ctg\alpha=\frac{1}{tg\alpha}=-\frac{12}{5}=-2,4\\\\5)tg4x=\sqrt{3}\\\\4x=arctg\sqrt{3}+\pi n,n\in Z\\\\4x=\frac{\pi }{3}+pi n,n\in Z\\\\x=\frac{\pi }{12}+\frac{\pi n }{4},n\in Z$

$6)Cos2x+\sqrt{3}Cosx+1=0\\\\2Cos^{2}x-1+\sqrt{3}Cosx+1=0\\\\2Cos^{2}x+\sqrt{3}Cosx=0\\\\Cosx(2Cosx+\sqrt{3})=0\\\\1)Cosx=0\\\\x=\frac{\pi }{2}+\pi n,n\in Z\\\\2)2Cosx+\sqrt{3}=0\\\\Cosx=-\frac{\sqrt{3}}{2}\\\\x= \pm arcCos(-\frac{\sqrt{3}} {2})+2\pi n,n \in Z \\\\x=\pm \frac{5\pi }{6}+2\pi n,n\in Z$

$7)Sin\frac{x}{2}\geq -\frac{\sqrt{2}}{2} \\\\-\frac{\pi }{4}+2\pi n\leq \frac{x}{2} \leq \frac{5\pi }{4}+2\pi n, n\in Z\\\\-\frac{\pi }{2}+4\pi n\leq x\leq \frac{5\pi }{2}+4\pi n,n\in Z\\\\Otvet:x\in[-\frac{\pi }{2}+4\pi n;\frac{5\pi }{2}+4\pi n],n\in Z$