Условие задачи
a) Решите уравнение:
\(cos7x+cos3x+2sin^2x=1\)
б) Найдите все корни уравнения на отрезке \(\displaystyle [0;\frac{\pi}{2}].\)
Решение
а) преобразуем сумму \(cos 7x + cos 3x\) в произведение и запишем \(1 - 2sin ^2 x = cos2x\)
\(2cos 5x cos 2x = cos 2x\)
\((2cos 5x-1)cos2x = 0\)
\(\displaystyle \left[ \begin{gathered} \cos 2x =0\\ \cos 5x = \frac{1}{2} \\ \end{gathered} \right. \)
\(\displaystyle \left[ \begin{gathered} 2x = \frac{\pi}{2}+\pi n ,\, n \in Z\\ 5x = \pm \frac{\pi}{3} +2\pi k, \, k \in Z\\ \end{gathered} \right.\)
\(\displaystyle \left[ \begin{gathered} x=\frac{\pi}{4}+\frac{\pi n}{2} ,\, n \in Z\\ x=\pm \frac{\pi}{15}+\frac{2\pi k}{5}, \, k \in Z\\ \end{gathered} \right.\)
б) Сделаем отбор корней с помощью двойного неравенства
1) \(\displaystyle 0 \leq \frac{\pi}{4} + \pi n \leq \frac{\pi}{2}\)
\(\displaystyle 0 \leq \frac{\pi}{4} + \pi n \leq \frac{\pi}{2}\)
\(\displaystyle 0 \leq \frac{1}{2} + n \leq 1\)
\(\displaystyle - \frac{1}{2} \leq n \leq \frac{1}{2}\)
\(n \in Z,\)
\(n = 0\)
\(\displaystyle x = \frac{\pi}{4}\)
2) \(\displaystyle 0 \leq \frac{\pi}{15} + \frac{2 \pi k}{5} \leq \frac{\pi}{2}\)
Отсюда \(\displaystyle - \frac{1}{6} \leq k \leq \frac{13}{12}, k=0\) или \(k=1,\)
\(\displaystyle x = \frac{\pi}{15}\) или \(\displaystyle x = \frac{7 \pi}{15}\)
3) \(\displaystyle 0 \leq - \frac{\pi}{15} + \frac{2 \pi k}{5} \leq \frac{\pi}{2}\)
\(\displaystyle \frac{1}{6} \leq k \leq \frac{17}{12}, \, \, k=1,\)
\(\displaystyle x = - \frac{\pi}{15} + \frac{2 \pi}{5} = \frac{\pi}{3}\)
Ответ
а) \(\displaystyle \left[ \begin{gathered} x=\frac{\pi}{4}+\frac{\pi k}{2} \\ x=\pm \frac{\pi}{15}+\frac{2\pi n}{5} \\ \end{gathered} \right.\; \; \; k, n \in Z\)
б) \(\displaystyle \frac{\pi}{4}; \frac{\pi}{15}; \frac{\pi}{3}; \frac{7 \pi}{15}\)
