Question

Тригонометрия. Задание в прикрепленном файле

Answer 1

$sin2x= sqrt{3}cos2x \ sin2x- sqrt{3}cos2x =0 (cdot frac{1}{2} ) \ cos frac{pi}{3}sin2x - sin frac{pi}{3}cos2x = 0 \ sin(2x- frac{pi}{3})=0 \ 2x= pi k + frac{pi}{3} \ x = frac{pi k}{2} + frac{pi}{6} \ boxed{k in Z}$
Дан промежуток: $x in [-pi;0]$
$k = 0 ; x = frac{pi}{6} approx frac{3,14}{6} notin [-pi;0] \ k = -1 ; x = -frac{2pi}{6} approx -frac{6,28}{6} in [-pi;0] \ k=-2 ; x = -frac{5 pi}{6} approx -frac{15,7}{6} in [-pi;0] \ k = -3 ; x = -frac{8pi}{6} approx -frac{21,98}{6} notin [-pi;0]$
Ответ: $-frac{2pi}{6} ; -frac{5 pi}{6}$