Помогите пожалуйста! 1)arcsin √3/2 +arcctg 1 + arccos (-1) 2)cos(2 arcsin 1/2)...

Question 1

Помогите пожалуйста!
1)arcsin √3/2 +arcctg 1 + arccos (-1)
2)cos(2 arcsin 1/2)
3)sin^x-4sinx cosx + 3cos^x=0
4)cosx<-√3/2<br> 5)sin2x≥1/2

Question 2

Решите задачу:

$1)\; \; arcsin\frac{\sqrt3}{2}+arcctg1 +arccos(-1)=\frac{\pi}{3}+\frac{\pi}{4}+\pi =\frac{19\pi }{12}\\\\2)\; \; cos(2arcsin \frac{1}{2})=cos(2 \alpha )\; ,\; \; \alpha =arcsin \frac{1}{2} \; ;\\\\cos2 \alpha =cos^2 \alpha -sin^2 \alpha \\\\sin(arcsin \alpha )= \alpha \; ,\; \; -1\leq \alpha \leq 1\\\\cos \alpha =\sqrt{1-sin^2 \alpha }\; \; \to \; \; cos(arcsin \frac{1}{2})=\sqrt{1-sin^2(arcsin\frac{1}{2})}=\\\\=\sqrt{1-(\frac{1}{2})^2}=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt3}{2}\\\\cos(2arcsin \frac{1}{2})= cos^2(arcsin\frac{1}{2})-sin^2(arcsin\frac{1}{2})=\\\\=(\frac{\sqrt3}{2})^2-(\frac{1}{2} )^2=\frac{3}{4}- \frac{1}{4}= \frac{2}{4}=\frac{1}{2}$

$3)\; \; sin^2x-4sinx\cdot cosx+3cos^2x=0\; |:cos^2x\ne 0\\\\tg^2x-4tgx+3=0\\\\tgx=1\; \; \; ili\; \; \; tgx=3\; \; (teorema\; Vieta)\\\\x=\frac{\pi}{4}+\pi n,\; n\in Z\; \; ili\; \; \; x=arctg3+\pi k,\; k\in Z$

$4)\; \; cosx\ \textless \ -\frac{\sqrt3}{2}\\\\\frac{5\pi }{6}+2\pi n\ \textless \ x\ \textless \ \frac{7\pi}{6}+2\pi n\; ,\; \; n\in Z\\\\5)\; \; sin2x \geq \frac{1}{2}\\\\ \frac{\pi}{6}+2\pi n\leq 2x \leq \frac{5\pi}{6}+2\pi \; ,\; n\in Z\\\\ \frac{\pi }{12}+\pi n \leq x\leq \frac{5\pi }{12}+\pi n\; ,\; n\in Z$