Преобразуем уравнение:
4 sin 2 x = sin x cos x ⇔ 4 sin 2 x − sin x cos x = 0 4\sin^2 x = \frac{\sin x}{\cos x} \Leftrightarrow 4\sin^2 x - \frac{\sin x}{\cos x} = 0 4 sin 2 x = cos x sin x ⇔ 4 sin 2 x − cos x sin x = 0 sin x ( 4 sin x cos x − 1 ) cos x = 0 \frac{\sin x (4\sin x \cos x - 1)}{\cos x} = 0 cos x sin x ( 4 sin x cos x − 1 ) = 0 Дробь равна 0 0 0 0 0 0 0 0 0 { [ sin x = 0 4 sin x cos x = 1 cos x ≠ 0 ⇔ { [ sin x = 0 sin 2 x = 1 2 cos x ≠ 0 ⇔ \begin{cases}
\left[\begin{array}{ll}
\sin x = 0 \\
4\sin x \cos x = 1
\end{array}
\right. \\
\cos x \neq 0
\end{cases} \Leftrightarrow \begin{cases}
\left[\begin{array}{ll}
\sin x = 0 \\
\sin 2x = \frac{1}{2}
\end{array}
\right. \\
\cos x \neq 0
\end{cases} \Leftrightarrow ⎩ ⎨ ⎧ [ sin x = 0 4 sin x cos x = 1 cos x = 0 ⇔ ⎩ ⎨ ⎧ [ sin x = 0 sin 2 x = 2 1 cos x = 0 ⇔ ⇔ { [ x = π k 2 x = π 6 + 2 π k 2 x = 5 π 6 + 2 π k x ≠ π 2 + π k , k ∈ Z ⇔ { [ x = π k x = π 12 + π k x = 5 π 12 + π k x ≠ π 2 + π k , k ∈ Z \Leftrightarrow\begin{cases}
\left[\begin{array}{ll}
x = \pi k\\
2x = \tfrac{\pi}{6} + 2\pi k\\
2x = \tfrac{5\pi}{6} + 2\pi k
\end{array}
\right. \\
x \neq \frac{\pi}{2} + \pi k
\end{cases}, k \in Z \Leftrightarrow \begin{cases}
\left[\begin{array}{ll}
x = \pi k\\
x = \tfrac{\pi}{12} + \pi k\\
x = \tfrac{5\pi}{12} + \pi k
\end{array}
\right. \\
x \neq \frac{\pi}{2} + \pi k
\end{cases}, k \in Z ⇔ ⎩ ⎨ ⎧ x = π k 2 x = 6 π + 2 π k 2 x = 6 5 π + 2 π k x = 2 π + π k , k ∈ Z ⇔ ⎩ ⎨ ⎧ x = π k x = 12 π + π k x = 12 5 π + π k x = 2 π + π k , k ∈ Z
б) Отберём корни на отрезке [ − π ; 0 ] [-\pi; 0] [ − π ; 0 ]
Корни, принадлежащие отрезку:
− π ; − 11 π 12 ; − 7 π 12 ; 0 -\pi; -\frac{11\pi}{12}; -\frac{7\pi}{12}; 0 − π ; − 12 11 π ; − 12 7 π ; 0
Ответ: а) π k ; π 12 + π k ; 5 π 12 + π k ; k ∈ Z \pi k;\frac{\pi}{12} + \pi k;\frac{5\pi}{12} + \pi k; k \in \mathbb{Z} π k ; 12 π + π k ; 12 5 π + π k ; k ∈ Z
б) − π ; − 11 π 12 ; − 7 π 12 ; 0. -\pi; -\frac{11\pi}{12}; -\frac{7\pi}{12}; 0. − π ; − 12 11 π ; − 12 7 π ; 0.
