Question

Решите пожалуйста... (10 класс)
$ctg 3x= sqrt{3}$
$sin (3-2x)=- frac{ sqrt{2} }{2}$
$cos pi sqrt{x} =- frac{ sqrt{3} }{2}$

Answer 1

1.
$ctg 3x= sqrt{3} \ 3x = frac{pi}{6} + 2pi cdot n : n in Z \ x = frac{pi}{18} + frac{2pi}{3} cdot n : n in Z$


2.
$sin (3-2x)=- frac{ sqrt{2} }{2} \ 3-2x1 = - frac{pi}{4} + 2pi n \ 3-2x2 = frac{5pi}{4} + 2pi n \ - 2x1 = - frac{pi}{4} - 3 + 2pi n \ - 2x2 = frac{5pi}{4} - 3 + 2pi n \ \ x1 = frac{pi}{8} + frac{3}{2} - pi n \ x2 = - frac{5pi}{8}+ frac{3}{2} - pi n$

3.
$cos(pi sqrt{x}) =- frac{ sqrt{3} }{2} \ pi sqrt{x} = pmfrac{5pi}{6} + 2pi n \ sqrt{x} = pmfrac{5}{6} + 2n \ x = (pmfrac{5}{6} + 2n)^{2} = frac{25}{36} + 4n^{2} pm{1frac{2}{3}n}$

Answer 2

Спасибо)