Question

1)arccos(-1)-arccos 1/2 -3 arccos(- √3/2)

2)2сos t=1

3)-2cos t=0

4)cos (arcctg √3)

5)arcctg (ctg 2П/3)

6)ctg x=-0,5

7)ctg x=0

Answer 1

============== 1 ==============

$arccos(-1)-arccos frac{1}{2} -3 arccos(- frac{ sqrt{3} }{2} )= \ = pi -arccos1-arccos frac{1}{2}-3( pi -arccos frac{ sqrt{3} }{2}) = pi -0- frac{ pi }{3} -3( pi - frac{ pi }{6})=\ = frac{2 pi }{3}- frac{5 pi }{2} =frac{4 pi }{6}- frac{15 pi }{6} =- frac{ 11pi }{6}$

============== 2 ==============

$2cos t=1\ cost= frac{1}{2} \ t=pm arccosfrac{1}{2}+2 pi k, k in Z\ t=pm frac{ pi }{3}+2 pi k, k in Z$

============== 3 ==============

$-2cos t=0 |:(-2)\ cost=0\ t= frac{ pi }{2} + pi k, k in Z$

============== 4 ==============

$cos (arcctg sqrt{3} )=cos frac{ pi }{6} = frac{ sqrt{3} }{2}$

============== 5 ==============

$arcctg (ctg frac{2 pi }{3} )= frac{2 pi }{3}$

============== 6 ==============

$ctg x=-0,5\ x=arcctg(-0.5)+ pi n, n in Z\ x= pi -arcctg(0.5)+ pi n, n in Z$

============== 7 ==============

$ctg x=0\ x=arcctg0+ pi n, n in Z\ x= frac{ pi }{2} + pi n, n in Z$