Условие задачи
а) Решить уравнение \(\displaystyle \frac{2-2 sin^2x- \sqrt3 cosx }{ctgx- \sqrt3} =0\)
б) найти все его корни на отрезке \([- \pi ;3 \pi ]\)
Решение
\(\displaystyle \frac{2-2 sin^2x- \sqrt3 cosx }{ctgx- \sqrt3} =0 \Leftrightarrow \left\{\begin{gathered}2(1-sin^2x )- \sqrt3 cosx=0 \\ ctgx \ne \sqrt3 \\ sinx \ne 0 \end{gathered}\right. \Leftrightarrow \left\{\begin{gathered}2cosx^2- \sqrt3 cosx=0 \\ x \ne \frac{\pi}{6}+ \pi n \\ x\ne \pi k \end{gathered}\right.\)
\(\displaystyle \Leftrightarrow \left\{\begin{gathered}cosx(2cosx-\sqrt{3})=0 \\ x\ne \frac{\pi}{6} +\pi n \\x\ne \pi k \end{gathered}\right. \Leftrightarrow \left\{\begin{gathered}\left[ \begin{array}{ccc} cos x=0 \\ cosx=\frac{\sqrt3}{2} \\ \end{array} \right. \\ x \ne \frac{\pi}{6}+\pi n \\ x\ne \pi k \end{gathered}\right.\Leftrightarrow \left\{\begin{gathered}\left[ \begin{array}{ccc} x=\frac{\pi}{2}+\pi n \\ x=\pm \frac{\pi}{6}+2\pi k \\ \end{array} \right. \\ x \ne \frac{\pi}{6}+\pi n \\ x\ne \pi k \end{gathered}\right. \Leftrightarrow\)
\(\displaystyle \left[ \begin{gathered} x= \frac{\pi}{2}+\pi n \\ x=-\frac{\pi}{6}+2\pi k \\ \end{gathered} \right. \, \, \, n,k \in Z\)
б) Отбор решений с помощью двойного неравенства
Найдем решение для серии \(\displaystyle x= \frac{\pi} {2}+ \pi n \)
\(\displaystyle - \pi \leq \frac{\pi} {2} + \pi n \leq 3 \pi\)
Решая это неравенство, получаем, что \(\displaystyle \left[ \begin{gathered} n=-1 \\ n=0 \\n=1\\n=2 \end{gathered} \right. \Rightarrow \left[ \begin{gathered} x= -\frac{\pi}{2} \\ x= \frac{\pi}{2}\\x= \frac{3\pi}{2}\\x= \frac{5\pi}{2} \end{gathered} \right.\)
Найдем решение серии \(\displaystyle x=- \frac{\pi} {6}+2 \pi k\)
\(\displaystyle - \pi \leq - \frac{\pi} {6} +2 \pi k \leq 3 \pi \Rightarrow \left[ \begin{gathered} k=0 \\ k=1 \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered} x = - \frac{\pi}{6} \\ x = \frac{11\pi}{6} \\ \end{gathered} \right.\)
Ответ
а) \(\displaystyle \left[ \begin{gathered} x= \frac{\pi}{2}+\pi n \\ x=-\frac{\pi}{6}+2\pi k \\ \end{gathered} \right. \, \, \, n,k \in Z\)
б) \(\displaystyle -\frac{\pi}{2}; \, -\frac{\pi}{6}; \, \frac{\pi}{2}; \, \frac{3\pi}{2}; \, \frac{11\pi}{6}; \, \frac{5\pi}{2}.\)