Question

Помогите с тригонометрией

Answer 1

$sinleft(dfrac{pi}{3} - 2x right) = -2cos^{2}left(dfrac{pi}{12} + x right) - 1$

$sindfrac{pi}{3} cos 2x - cosdfrac{pi}{3}sin 2x = -left(2cos^{2}left(dfrac{pi}{12} + x right) - 1 + 2right)$

$dfrac{sqrt{3}}{2}cos 2x - dfrac{1}{2}sin 2x = -left(cos left(dfrac{pi}{6} + 2x right) + 2 right)$

$dfrac{sqrt{3}}{2}cos 2x - dfrac{1}{2}sin 2x = -cosleft(dfrac{pi}{6} + 2x right) - 2$

$dfrac{sqrt{3}}{2}cos 2x - dfrac{1}{2}sin 2x = sin dfrac{pi}{6}sin 2x - cos dfrac{pi}{6}cos 2x - 2$

$dfrac{sqrt{3}}{2}cos 2x - dfrac{1}{2}sin 2x = dfrac{1}{2} sin 2x - dfrac{sqrt{3}}{2}cos 2x - 2$

$sqrt{3}cos 2x - sin 2x = -2 | : dfrac{1}{sqrt{(sqrt{3})^{2} + (-1)^{2}}}$

$dfrac{sqrt{3}}{2}cos 2x - dfrac{1}{2} sin 2x = -1$

$cos dfrac{pi}{6} cos 2x - sindfrac{pi}{6}sin 2x = -1$

$cos left(dfrac{pi}{6} + 2x right) = -1$

$dfrac{pi}{6} + 2x = pi + 2pi n, n in Z$

$2x = pi - dfrac{pi}{6} + 2pi n, n in Z$

$2x = dfrac{5pi}{6} + 2pi n, n in Z$

$x = dfrac{5pi}{12} + pi n, n in Z$

Ответ: $x = dfrac{5pi}{12} + pi n, n in Z$

Answer 2

Каких-либо объяснений здесь нет. Если будет что-то не понятно,— спрашивайте.