Решение:
а) Распишем косинус по формуле двойного угла
cos 2 x = 2 cos 2 x − 1 \cos 2x = 2\cos^2 x - 1 cos 2 x = 2 cos 2 x − 1
Тогда уравнение примет вид
4 cos 3 x − 2 3 ( 2 cos 2 x − 1 ) + 3 cos x = 2 3 4\cos^3 x - 2\sqrt{3}\,(2\cos^2 x - 1) + 3\cos x = 2\sqrt{3} 4 cos 3 x − 2 3 ( 2 cos 2 x − 1 ) + 3 cos x = 2 3 4 cos 3 x − 4 3 cos 2 x + 2 3 + 3 cos x = 2 3 4\cos^3 x - 4\sqrt{3}\cos^2 x + 2\sqrt{3} + 3\cos x = 2\sqrt{3} 4 cos 3 x − 4 3 cos 2 x + 2 3 + 3 cos x = 2 3 4 cos 3 x − 4 3 cos 2 x + 3 cos x = 0 4\cos^3 x - 4\sqrt{3}\cos^2 x + 3\cos x = 0 4 cos 3 x − 4 3 cos 2 x + 3 cos x = 0 cos x ( 4 cos 2 x − 4 3 cos x + 3 ) = 0 \cos x\,(4\cos^2 x - 4\sqrt{3}\cos x + 3) = 0 cos x ( 4 cos 2 x − 4 3 cos x + 3 ) = 0 cos x ( 2 cos x − 3 ) 2 = 0 \cos x\,(2\cos x - \sqrt{3})^2 = 0 cos x ( 2 cos x − 3 ) 2 = 0 { cos x = 0 2 cos x − 3 = 0 ⟺ { cos x = 0 cos x = 3 2 \left\{
\begin{array}{l}
\cos x = 0\\
2\cos x - \sqrt{3} = 0
\end{array}
\right.
\Longleftrightarrow
\left\{
\begin{array}{l}
\cos x = 0\\
\cos x = \dfrac{\sqrt{3}}{2}
\end{array}
\right. { cos x = 0 2 cos x − 3 = 0 ⟺ ⎩ ⎨ ⎧ cos x = 0 cos x = 2 3 [ x = π 2 + π k , k ∈ Z x = π 6 + 2 π k , k ∈ Z x = 11 π 6 + 2 π k , k ∈ Z \left[
\begin{array}{l}
x = \dfrac{\pi}{2} + \pi k,\; k \in Z\\
x = \dfrac{\pi}{6} + 2\pi k,\; k \in Z\\
x = \dfrac{11\pi}{6} + 2\pi k,\; k \in Z
\end{array}
\right. x = 2 π + π k , k ∈ Z x = 6 π + 2 π k , k ∈ Z x = 6 11 π + 2 π k , k ∈ Z
б) Отберем подходящие корни с помощью неравенств.
x = π 2 + π k x = \dfrac{\pi}{2} + \pi k x = 2 π + π k 2 π ≤ π 2 + π k ≤ 7 π 2 ⟺ 2 ≤ 1 2 + k ≤ 7 2 ⟺ 3 2 ≤ k ≤ 6 2 2\pi \le \dfrac{\pi}{2} + \pi k \le \dfrac{7\pi}{2}
\Longleftrightarrow
2 \le \dfrac{1}{2} + k \le \dfrac{7}{2}
\Longleftrightarrow
\dfrac{3}{2} \le k \le \dfrac{6}{2} 2 π ≤ 2 π + π k ≤ 2 7 π ⟺ 2 ≤ 2 1 + k ≤ 2 7 ⟺ 2 3 ≤ k ≤ 2 6 [ k = 2 k = 3 ⇒ [ x = π 2 + 2 π x = π 2 + 3 π ⇒ [ x = 5 π 2 x = 7 π 2 \left[
\begin{array}{l}
k = 2\\
k = 3
\end{array}
\right.
\Rightarrow
\left[
\begin{array}{l}
x = \dfrac{\pi}{2} + 2\pi\\
x = \dfrac{\pi}{2} + 3\pi
\end{array}
\right.
\Rightarrow
\left[
\begin{array}{l}
x = \dfrac{5\pi}{2}\\
x = \dfrac{7\pi}{2}
\end{array}
\right. [ k = 2 k = 3 ⇒ x = 2 π + 2 π x = 2 π + 3 π ⇒ x = 2 5 π x = 2 7 π
x = π 6 + 2 π k x = \dfrac{\pi}{6} + 2\pi k x = 6 π + 2 π k 2 π ≤ π 6 + 2 π k ≤ 7 π 2 ⟺ 2 ≤ 1 6 + 2 k ≤ 7 2 ⟺ 11 6 ≤ 2 k ≤ 20 6 2\pi \le \dfrac{\pi}{6} + 2\pi k \le \dfrac{7\pi}{2}
\Longleftrightarrow
2 \le \dfrac{1}{6} + 2k \le \dfrac{7}{2}
\Longleftrightarrow
\dfrac{11}{6} \le 2k \le \dfrac{20}{6} 2 π ≤ 6 π + 2 π k ≤ 2 7 π ⟺ 2 ≤ 6 1 + 2 k ≤ 2 7 ⟺ 6 11 ≤ 2 k ≤ 6 20 11 12 ≤ k ≤ 10 12 ⇒ k = 1 ⇒ x = π 6 + 2 π = 13 π 6 \dfrac{11}{12} \le k \le \dfrac{10}{12}
\Rightarrow
k = 1
\Rightarrow
x = \dfrac{\pi}{6} + 2\pi = \dfrac{13\pi}{6} 12 11 ≤ k ≤ 12 10 ⇒ k = 1 ⇒ x = 6 π + 2 π = 6 13 π
x = 11 π 6 + 2 π k x = \dfrac{11\pi}{6} + 2\pi k x = 6 11 π + 2 π k 2 π ≤ 11 π 6 + 2 π k ≤ 7 π 2 ⟺ 2 ≤ 11 6 + 2 k ≤ 7 2 2\pi \le \dfrac{11\pi}{6} + 2\pi k \le \dfrac{7\pi}{2}
\Longleftrightarrow
2 \le \dfrac{11}{6} + 2k \le \dfrac{7}{2} 2 π ≤ 6 11 π + 2 π k ≤ 2 7 π ⟺ 2 ≤ 6 11 + 2 k ≤ 2 7 1 6 ≤ 2 k ≤ 10 6 ⟺ 1 12 ≤ k ≤ 10 12 ⇒ k ∈ ∅ \dfrac{1}{6} \le 2k \le \dfrac{10}{6}
\Longleftrightarrow
\dfrac{1}{12} \le k \le \dfrac{10}{12}
\Rightarrow
k \in \varnothing 6 1 ≤ 2 k ≤ 6 10 ⟺ 12 1 ≤ k ≤ 12 10 ⇒ k ∈ ∅
Ответ:
а) π 6 + 2 π k ; π 2 + π k ; 11 π 6 + 2 π k , k ∈ Z \dfrac{\pi}{6} + 2\pi k;\; \dfrac{\pi}{2} + \pi k;\; \dfrac{11\pi}{6} + 2\pi k,\; k \in Z 6 π + 2 π k ; 2 π + π k ; 6 11 π + 2 π k , k ∈ Z
б) 5 π 2 ; 13 π 6 ; 7 π 2 \dfrac{5\pi}{2};\; \dfrac{13\pi}{6};\; \dfrac{7\pi}{2} 2 5 π ; 6 13 π ; 2 7 π