Giải một vài bài lượng giác

Đề bài số 1

\(sin{x} + \sqrt{2-sin^2x}=2+\sqrt{1+cos{4x}}\)

Lời giải

Bài này sau khi em làm một hồi mà suy nghĩ nát óc chưa ra, thấy có căn, rồi mấy cái số 2 lặp đi lặp lại nữa rất nghi ngờ nó là p.trình lượng giác không mẫu mực, đánh giá 2 vế để có dấu bằng xảy ra. Cụ thể thế này:

Dùng B.C.S ta có: VT= \( 1.sinx+1.\sqrt{2-sin^2x} \leq \sqrt{(1^2+1^2)(sin^2x+2-sin^2x)} = 2 \)

Trong khi VP = \(2+\sqrt{1+cos4x} \geq 2\)

nên phương trình chỉ xảy ra khi cả 2 cùng thỏa dấu bằng xảy ra, nghĩa là:

\(\left\{ \begin{array}{l}\sin x = \sqrt {2 – {{\sin }^2}x} \\1 + \cos 4x = 0\end{array} \right.\)

Từ đó ta giải được \(\left\{ \begin{array}{l}\sin x = 1 \\cos4x =-1\end{array} \right.\)

Tức là vô nghiệm


Giải bài tập đề cương thầy Đức, phần 1

\[\begin{array}{l}1.1){\cos ^2}x – 2\sin x + 2 = 0\\ \Leftrightarrow  – {\sin ^2}x – 2\sin x + 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 1(n)\\\sin x =  – 3(l)\end{array} \right.\\ \Leftrightarrow x = \frac{\pi }{2} + k2\pi (k \in \mathbb{Z})\end{array}\]

\[\begin{array}{l}1.2)4\cos \frac{{5x}}{2}\cos \frac{{3x}}{2} + 2(8\sin x – 1)\cos x = 5\\ \Leftrightarrow 2(\cos 4x + \cos x) + 2(8\sin x\cos x – \cos x) = 5\\ \Leftrightarrow \cos 4x + \cos x + 4\sin 2x – \cos x = \frac{5}{2}\\ \Leftrightarrow 1 – 2{\sin ^2}2x + 4\sin 2x = \frac{5}{2}\\ \Leftrightarrow  – 2{\sin ^2}2x + 4\sin 2x – \frac{3}{2} = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \frac{3}{2}(l)\\\sin x = \frac{1}{2}(n)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.(k \in \mathbb{Z})\end{array}\]

\[\begin{array}{l}1.3)\cos 4x + 12{\sin ^2}x – 1 = 0\\ \Leftrightarrow 1 – 2{\sin ^2}2x + 12{\sin ^2}x – 1 = 0\\ \Leftrightarrow 6{\sin ^2}x – {\sin ^2}2x = 0\\ \Leftrightarrow 6{\sin ^2}x – 4{\sin ^2}x{\cos ^2}x = 0\\ \Leftrightarrow 2{\sin ^2}x(3 – 2{\cos ^2}x) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\{\cos ^2}x = \frac{3}{2}(l)\end{array} \right.\\ \Leftrightarrow x = k\pi (k \in \mathbb{Z})\end{array}\]

\[\begin{array}{l}1.4)\frac{{2({{\cos }^6}x + {{\sin }^6}x) – \sin x\cos x}}{{\sqrt 2  – 2\cos x}} = 0(1)\\DK:\sqrt 2  – 2\cos x \ne 0\\ \Leftrightarrow \cos x \ne \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow x \ne  \pm \frac{\pi }{4} + k2\pi (k \in \mathbb{Z})\\(1) \Leftrightarrow 2 – \frac{3}{2}{\sin ^2}2x – \frac{1}{2}\sin 2x = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = 1(n)\\\sin 2x = \frac{{ – 4}}{3}(l)\end{array} \right.\\ \Leftrightarrow 2x = \frac{\pi }{2} + k2\pi \\ \Leftrightarrow x = \frac{\pi }{4} + k\pi (k \in Z)\\So {\rm{dk:}} \Leftrightarrow x = \frac{{5\pi }}{4} + k\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.5)4{\sin ^3}x + 4{\sin ^2}x + 3\sin 2x + 6\cos x = 0\\ \Leftrightarrow 4{\sin ^2}x(\sin x + 1) + 6\cos x(\sin x + 1) = 0\\ \Leftrightarrow (\sin x + 1)(4{\sin ^2}x + 6\cos x) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x =  – 1(1)\\4{\sin ^2}x + 6\cos x = 0(2)\end{array} \right.\\(1) \Leftrightarrow x = \frac{{ – \pi }}{2} + k2\pi (k \in Z)\\(2) \Leftrightarrow 4 – 4{\cos ^2}x + 6\cos x = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 2(l)\\\cos x = \frac{{ – 1}}{2}(n)\end{array} \right.\\ \Leftrightarrow x =  \pm \frac{{2\pi }}{3} + k2\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.6)5\left( {\sin x + \frac{{\cos 3x + \sin 3x}}{{1 + 2\sin 2x}}} \right) = \cos 2x + 3(1)\\DK:1 + 2\sin 2x \ne 0\\ \Leftrightarrow \sin 2x \ne \frac{{ – 1}}{2}\\ \Leftrightarrow \left[ \begin{array}{l}x \ne \frac{{ – \pi }}{{12}} + k\pi \\x \ne \frac{{7\pi }}{{12}} + k\pi \end{array} \right.(k \in Z)\\(1) \Leftrightarrow 5(\sin x + 2\sin x\sin 2x + \cos 3x + \sin 3x) = (\cos 2x + 3)(1 + 2\sin 2x)\\ \Leftrightarrow 5(\sin x – \cos 3x + \cos x + \cos 3x + \sin 3x) = (\cos 2x + 3)(1 + 2\sin 2x)\\ \Leftrightarrow 5(\sin x + \sin 3x + \cos x) = (\cos 2x + 3)(1 + 2\sin 2x)\\ \Leftrightarrow 5(2\sin 2x\cos x + \cos x) = (\cos 2x + 3)(1 + 2\sin 2x)\\ \Leftrightarrow 5\cos x(2\sin 2x + 1) = (\cos 2x + 3)(1 + 2\sin 2x)\\ \Leftrightarrow 5\cos x = \cos 2x + 3\\ \Leftrightarrow 2{\cos ^2}x – 5\cos x + 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 2(l)\\\cos x = \frac{1}{2}(n)\end{array} \right.\\ \Leftrightarrow x =  \pm \frac{\pi }{3} + k2\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.7)\cot x – \tan x + 4\sin 2x = \frac{2}{{\sin 2x}}(1)\\DK:\left\{ \begin{array}{l}\sin x \ne 0\\\cos x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne k\pi \\x \ne \frac{\pi }{2} + k\pi \end{array} \right.(k \in Z)\\(1) \Leftrightarrow 2\frac{{\cos 2x}}{{\sin 2x}} + 4\sin 2x = \frac{2}{{\sin 2x}}\\ \Leftrightarrow \cos 2x + 2{\sin ^2}2x = 1\\ \Leftrightarrow \cos 2x + 1 – 2{\cos ^2}2x = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = 1\\\cos 2x = \frac{{ – 1}}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x =  \pm \frac{\pi }{3} + k\pi \end{array} \right.(k \in Z)\\So dk: \Leftrightarrow x =  \pm \frac{\pi }{3} + k\pi (k \in \mathbb{Z})\end{array}\]

\[\begin{array}{l}1.8)5\sin x – 2 = 3(1 – \sin x){\tan ^2}x(1)\\DK:\cos x \ne 0 \Leftrightarrow x \ne \frac{\pi }{2} + k\pi (k \in Z)\\(1) \Leftrightarrow 5\sin x – 2 = 3(1 – \sin x)\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}\\ \Leftrightarrow 5\sin x – 2 = 3(1 – \sin x)\frac{{{{\sin }^2}x}}{{(1 + \sin x)(1 – \sin x)}}\\ \Leftrightarrow 5\sin x – 2 = \frac{{3{{\sin }^2}x}}{{1 + \sin x}}\\ \Leftrightarrow 5{\sin ^2}x + 3\sin x – 2 = 3{\sin ^2}x\\ \Leftrightarrow 2{\sin ^2}x + 3\sin x – 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x =  – 2(l)\\\sin x = \frac{1}{2}(n)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.(k \in Z)\\So dk : \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.(k \in Z)\end{array}\]

\[\begin{array}{l}1.9){\cos ^2}3x\cos 2x – {\cos ^2}x = 0\\ \Leftrightarrow \left( {1 + \cos 6x} \right)\cos 2x – 1 – \cos 2x = 0\\ \Leftrightarrow \cos 2x + \cos 6x\cos 2x – 1 – \cos 2x = 0\\ \Leftrightarrow \frac{{\cos 8x}}{2} + \frac{{\cos 4x}}{2} – 1 = 0\\ \Leftrightarrow 2{\cos ^2}4x + \cos 4x – 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos 4x = 1(n)\\\cos 4x = \frac{{ – 3}}{2}(l)\end{array} \right.\\ \Leftrightarrow 4x = k2\pi \\ \Leftrightarrow x = k\frac{\pi }{2}(k \in Z)\end{array}\]

\[\begin{array}{l}1.10){\cos ^4}x + {\sin ^4}x + \cos \left( {x – \frac{\pi }{4}} \right)\sin \left( {3x – \frac{\pi }{4}} \right) – \frac{3}{2} = 0\\ \Leftrightarrow  – \frac{1}{2} – \frac{1}{2}{\sin ^2}2x + \frac{1}{2}\sin \left( {4x – \frac{\pi }{2}} \right) + \frac{1}{2}\sin 2x = 0\\ \Leftrightarrow  – 1 – {\sin ^2}2x – \cos 4x + \sin 2x = 0\\ \Leftrightarrow {\sin ^2}2x – \sin 2x + 2{\cos ^2}2x = 0\\ \Leftrightarrow  – {\sin ^2}2x – \sin 2x + 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = 1(n)\\\sin 2x =  – 2(l)\end{array} \right.\\ \Leftrightarrow x = \frac{\pi }{4} + k\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.11)\cot x = \tan x + \frac{{2\cos 4x}}{{\sin 2x}}(1)\\DK:\left\{ \begin{array}{l}x \ne k\pi \\x \ne \frac{\pi }{2} + k\pi \end{array} \right.(k \in Z)\\(1) \Leftrightarrow \frac{{\cos x}}{{\sin x}} = \frac{{\sin x}}{{\cos x}} + \frac{{\cos 4x}}{{\sin x\cos x}}\\ \Leftrightarrow {\cos ^2}x = {\sin ^2}x + \cos 4x\\ \Leftrightarrow \cos 2x = \cos 4x\\ \Leftrightarrow \left[ \begin{array}{l}2x = 4x + k2\pi \\2x =  – 4x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  – k\pi \\x = k\frac{\pi }{3}\end{array} \right.(k \in Z)\\So dk: \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\pi \\x = \frac{{2\pi }}{3} + k\pi \end{array} \right.(k \in Z)\end{array}\]

\[\begin{array}{l}1.12)\frac{{\sin 2x}}{{\cos x}} + \frac{{\cos 2x}}{{\sin x}} = \tan x – \cot x(1)\\DK:\left\{ \begin{array}{l}x \ne k\pi \\x \ne \frac{\pi }{2} + k\pi \end{array} \right.(k \in Z)\\(1) \Leftrightarrow \sin 2x\sin x + \cos 2x\cos x = {\sin ^2}x – {\cos ^2}x\\ \Leftrightarrow \cos (2x – x) =  – \cos 2x\\ \Leftrightarrow \cos x + \cos 2x = 0\\ \Leftrightarrow 2{\cos ^2}x + \cos x – 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x =  – 1\\\cos x = \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \pi  + k2\pi \\x =  \pm \frac{\pi }{3} + k2\pi \end{array} \right.(k \in Z)\\So DK: \Leftrightarrow x =  \pm \frac{\pi }{3} + k2\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.13)3\sin x + \cos 2x + \sin 2x = 4\sin x{\cos ^2}\frac{x}{2}\\ \Leftrightarrow 3\sin x + \cos 2x + \sin 2x = 2\sin x + \sin 2x\\ \Leftrightarrow \sin x + \cos 2x = 0\\ \Leftrightarrow  – 2{\sin ^2}x + \sin x + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 1\\\sin x = \frac{{ – 1}}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x = \frac{{ – \pi }}{6} + k2\pi \\x = \frac{{7\pi }}{6} + k2\pi \end{array} \right.(k \in Z)\end{array}\]

\[\begin{array}{l}1.14)4({\sin ^4}x + {\cos ^4}x) + \cos 4x + \sin 2x = 0\\ \Leftrightarrow 4 – 2{\sin ^2}2x + \cos 4x + \sin 2x = 0\\ \Leftrightarrow  – 4{\sin ^2}2x + \sin 2x + 5 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 2x =  – 1(n)\\\sin 2x = \frac{5}{4}(l)\end{array} \right.\\ \Leftrightarrow 2x = \frac{{ – \pi }}{2} + k2\pi \\ \Leftrightarrow x = \frac{{ – \pi }}{4} + k\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.15)2{\cos ^2}x + 2\sqrt 3 \sin x\cos x + 1 = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\ \Leftrightarrow 3{\cos ^2}x + 2\sqrt 3 \sin x\cos x + {\sin ^2}x = 3\left( {\sin x + \sqrt 3 \cos x} \right)\\ \Leftrightarrow {\left( {\sin x + \sqrt 3 \cos x} \right)^2} – 3\left( {\sin x + \sqrt 3 \cos x} \right) = 0\\ \Leftrightarrow \left( {\sin x + \sqrt 3 \cos x} \right)\left( {\sin x + \sqrt 3 \cos x – 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin \left( {x + \frac{\pi }{3}} \right) = 0(n)\\\sin \left( {x + \frac{\pi }{3}} \right) = 3(l)\end{array} \right.\\ \Leftrightarrow x =  – \frac{\pi }{3} + k\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.16)\sin 2x + \cos 2x + 3\sin x – \cos x – 2 = 0\\ \Leftrightarrow  – 2{\sin ^2}x + \sin 2x – 1 + 2\sin x + \sin x – \cos x = 0\\ \Leftrightarrow  – 2\sin x(\sin x – \cos x – 1) + (\sin x – \cos x – 1) = 0\\ \Leftrightarrow (\sin x – \cos x – 1)\left( { – 2\sin x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x – \cos x = 1\\\sin x = \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{3\pi }}{4} + k2\pi \\x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.\end{array}\]

\[\begin{array}{l}1.17)\sin 2x + 2\cos 2x = 1 + \sin x – 4\cos x\\ \Leftrightarrow 2\sin x\cos x + 4{\cos ^2}x – 3 – \sin x + 6\cos x – 2\cos x = 0\\ \Leftrightarrow 3\left( {2\cos x – 1} \right) + 2\cos x\left( {2\cos x – 1} \right) + \sin x\left( {2\cos x – 1} \right) = 0\\ \Leftrightarrow \left( {2\cos x – 1} \right)\left( {3 + 2\cos x + \sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}(n)\\\sin x + 2\cos x =  – 3(l)\end{array} \right.\\ \Leftrightarrow x =  \pm \frac{\pi }{3} + k2\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.18)\cos 2x – {\tan ^2}x = \frac{{{{\cos }^2}x + {{\cos }^3}x – 1}}{{{{\cos }^2}x}}(1)\\DK:\cos x \ne 0 \Leftrightarrow x = \frac{\pi }{2} + k2\pi (k \in Z)\\(1)\cos 2x – {\tan ^2}x = 1 + \cos x – 1 – {\tan ^2}x\\ \Leftrightarrow \cos 2x = \cos x\\ \Leftrightarrow \left[ \begin{array}{l}2x = x + k2\pi \\2x =  – x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = k\frac{{2\pi }}{3}\end{array} \right.(k \in Z)\end{array}\]

\[\begin{array}{l}1.19){\sin ^2}\left( {x + \frac{\pi }{3}} \right) + {\sin ^2}\left( {x + \frac{{2\pi }}{3}} \right) = \frac{{3 – \sin x}}{2}  \\ \Leftrightarrow \cos \left( {2x + \frac{{2\pi }}{3}} \right) + \cos \left( {2x + \frac{{4\pi }}{3}} \right) = \sin x – 1\\ \Leftrightarrow \cos \left( {2x + \frac{{2\pi }}{3}} \right) – \cos \left( {2x + \frac{\pi }{3}} \right) = \sin x – 1\\ \Leftrightarrow \frac{{ – 1}}{2}\cos 2x – \frac{{\sqrt 3 }}{2}\sin 2x – \frac{1}{2}\cos 2x + \frac{{\sqrt 3 }}{2}\sin 2x = \sin x – 1\\ \Leftrightarrow \cos 2x = 1 – \sin x\\ \Leftrightarrow  – 2{\sin ^2}x + \sin x = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\sin x = \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{\pi }{6} + k2\pi \\x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.(k \in Z)\end{array}\]

\[\begin{array}{l}1.20)\cos 2x\left( {1 + \frac{1}{{\tan x\tan 2x + 1}}} \right) = {\cos ^2}x(1)\\DK\left\{ \begin{array}{l}\cos x \ne 0\\\cos 2x \ne 0\\\tan x\tan 2x \ne  – 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \frac{\pi }{2} + k\pi \\x \ne \frac{\pi }{4} + k\frac{\pi }{2}\\\tan x \ne \cot \left( { – 2x} \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \frac{\pi }{2} + k\pi \\x \ne \frac{\pi }{4} + k\frac{\pi }{2}\\x \ne  – \frac{\pi }{2} – k\pi \end{array} \right.(k \in Z)\\(1) \Leftrightarrow \cos 2x\left( {1 + \cos 2x} \right) = {\cos ^2}x\\ \Leftrightarrow \cos 2x + {\cos ^2}2x = {\cos ^2}x\\ \Leftrightarrow \cos 2x + 2{\cos ^2}2x – 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x =  – 1\\\cos 2x = \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x =  \pm \frac{\pi }{6} + k\pi \end{array} \right.(k \in Z)\\So dk; \Leftrightarrow x =  \pm \frac{\pi }{6} + k\pi (k \in Z)\end{array}\]

\[\begin{array}{l}1.21)\left( {1 – \tan x} \right)\left( {1 + \sin 2x} \right) = 1 + \tan x(1)\\DK:\cos x \ne 0 \Leftrightarrow x = \frac{\pi }{2} + k\pi \\(1) \Leftrightarrow 1 – \tan x + \sin 2x – \tan x\sin 2x = 1 + \tan x\\ \Leftrightarrow  – \tan x – {\sin ^2}x + \sin x\cos x = 0\\ \Leftrightarrow \sin x\left( {\frac{1}{{\cos x}} + \sin x – \cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\sin x\cos x – {\cos ^2}x + 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\\sin x\left( {\sin x + \cos x} \right) = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x =  – \frac{\pi }{4} + k\pi \end{array} \right.(k \in Z)\end{array}\]

\[\begin{array}{l}1.22)\frac{{\sqrt 3 }}{{{{\cos }^2}x}} + \frac{{4 + 2\sin 2x}}{{\sin 2x}} – 2\sqrt 3  = 2\left( {\cot x + 1} \right)(1)\\DK:\sin 2x \ne 0 \Leftrightarrow x \ne k\frac{\pi }{2}(k \in Z)\\(1) \Leftrightarrow 2\sqrt 3 \sin x + 4\cos x + 2\sin 2x\cos x – 2\sqrt 3 \sin 2x\cos x = 2\sin 2x\cos x + 2\cot x\sin 2x\cos x\\ \Leftrightarrow 2\sqrt 3 \sin x + 4\cos x – 2\sqrt 3 \sin 2x\cos x = 4{\cos ^3}x\\ \Leftrightarrow 2\sqrt 3 \sin x\left( {1 – 2{{\cos }^2}x} \right) + 4\cos x\left( {1 – {{\cos }^2}x} \right) = 0\\ \Leftrightarrow \sqrt 3 \sin x\cos 2x – 2\cos x{\sin ^2}x = 0\\ \Leftrightarrow \sin x\left( {\sqrt 3 \cos 2x – \sin 2x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\sqrt 3 \cos 2x – \sin 2x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\\cos \left( {2x + \frac{\pi }{6}} \right) = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{\pi }{6} + k\frac{\pi }{2}\end{array} \right.(k \in Z)\\So dk: \Leftrightarrow x = \frac{\pi }{6} + k\frac{\pi }{2}(k \in Z)\end{array}\]

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