Question

Вычислить:

2arcsin√3/2+3arcsin(-1/2)

arccos(-1/2)-arcsin√3

Решить:

cos(4-2x)=-1/2

√2cos(2х+п/4)+1=0

Answer 1

$1); ; 2arcsinfrac{sqrt3}{2}+3arcsin(-frac{1}{2})=2cdot frac{pi}{3}+3cdot (-frac{pi}{6})=-frac{pi}{6}\\2); ; arccos(-frac{1}{2})-arcsinsqrt3=(pi -frac{pi}{3})-arcsinsqrt3=frac{2pi}{3}-arcsinsqrt3\\3); ; cos(4-2x)=-frac{1}{2}\\4-2x=pm arccos(-frac{1}{2})+2pi n=pm (pi -frac{pi}{3})+2pi n=pm frac{2pi}{3}+2pi n\\-2x=pm frac{2pi}{3}-4+2pi n; ,; ; nin Z\\x=mpfrac{pi}{3}+2+pi n; ,; ; nin Z\\4); ; sqrt2cos(2x+frac{pi}{4})+1=0$

$cos(2x+frac{pi}{4})=-frac{sqrt2}{2}\\2x+frac{pi}{4}=pm arccos(-frac{sqrt2}{2})+2pi n=pm frac{3pi }{4}+2pi n,; nin Z\\2x=pm frac{3pi}{4}-frac{pi}{4}+2pi n; ,; nin Z\\x=pm frac{3pi }{8}-frac{pi}{8}+pi n; ,; nin Z$