Question

Помогите!! Дано: arcsin√3/2+arctg(tg6π/7)+arcctg(ctg(-π/3))=kπ найдите значение k​

Answer 1

Ответ:

$arcsin\dfrac{\sqrt3}{2}+arctg(tg\dfrac{6\pi }{7})-arcctg(ctg(-\dfrac{\pi}{3}))=k\pi ;\\\\\\arcsin\dfrac{\sqrt3}{2}+arctg(tg\dfrac{6\pi }{7})-arcctg(ctg(-\dfrac{\pi}{3}))=\dfrac{\pi}{3}+\dfrac{6\pi}{7}-arcctg(-ctg\dfrac{\pi}{3})=\\\\\\=\dfrac{\pi}{3}+\dfrac{6\pi}{7}-(\pi -\dfrac{\pi}{3})=\dfrac{\pi}{3}+\dfrac{6\pi}{7}-\dfrac{2\pi}{3}=-\dfrac{\pi}{3}+\dfrac{6\pi }{7}=\dfrac{-7\pi +18\pi }{21}=\dfrac{11\pi }{21}\ ;\\\\\\\dfrac{11\pi }{21}=k\pi \ \ \ \Rightarrow \ \ \ k=\dfrac{11}{21}$