Giải phương trình \(\cos x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2} - \sin x\sin \dfrac{x}{2}\sin \dfrac{{3x}}{2} = \dfrac{1}{2}\).
\(\begin{array}{l}\,\,\,\,\,\,\cos x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2} - \sin x\sin \dfrac{x}{2}\sin \dfrac{{3x}}{2} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{1}{2}\cos x\left( {\cos 2x + \cos x} \right) + \dfrac{1}{2}\sin x\left( {\cos 2x - \cos x} \right) = \dfrac{1}{2}\\ \Leftrightarrow \cos x\cos 2x + {\cos ^2}x + \sin x\cos 2x - \sin x\cos x = 1\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\cos x + {\cos ^2}x - 1 = 0\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\cos x - {\sin ^2}x = 0\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\left( {\sin x + \cos x} \right) = 0\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\cos 2x - \sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x + \cos x = 0\\\cos 2x - \sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = - \cos x\\1 - 2{\sin ^2}x - \sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\tan x = - 1\\\sin x = \dfrac{1}{2}\\\sin x = - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \dfrac{\pi }{4} + k\pi \\x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \\x = - \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy nghiệm của phương trình đã cho là: \(x = - \dfrac{\pi }{4} + k\pi ;\,\,x = \dfrac{\pi }{6} + k2\pi ;\)\(x = \dfrac{{5\pi }}{6} + k2\pi ;\,\,x = - \dfrac{\pi }{2} + k2\pi \)\(\left( {k \in \mathbb{Z}} \right)\).
Giải phương trình \(\cos 2x + \cos 4x + \cos 6x = \cos x\cos 2x\cos 3x + 2\).
\(\begin{array}{l}\,\,\,\,\,\,\cos 2x + \cos 4x + \cos 6x = \cos x\cos 2x\cos 3x + 2\\ \Leftrightarrow 2\cos 4x\cos 2x + \cos 4x = \dfrac{1}{2}\cos 2x\left( {\cos 4x + \cos 2x} \right) + 2\\ \Leftrightarrow 2\cos 4x\cos 2x + \cos 4x = \dfrac{1}{2}\cos 2x\cos 4x + \dfrac{1}{2}{\cos ^2}2x + 2\\ \Leftrightarrow \dfrac{3}{2}\cos 4x\cos 2x + \cos 4x = \dfrac{1}{2}{\cos ^2}2x + 2\\ \Leftrightarrow 3\cos 4x\cos 2x + 2\cos 4x = {\cos ^2}2x + 4\\ \Leftrightarrow 3\left( {2{{\cos }^2}2x - 1} \right)\cos 2x + 2\left( {2{{\cos }^2}2x - 1} \right) = {\cos ^2}2x + 4\\ \Leftrightarrow 6{\cos ^3}2x - 3\cos 2x + 4{\cos ^2}2x - 2 = {\cos ^2}2x + 4\\ \Leftrightarrow 6{\cos ^3}2x + 3{\cos ^2}2x - 3\cos 2x - 6 = 0\\ \Leftrightarrow 2{\cos ^3}2x + {\cos ^2}2x - \cos 2x - 2 = 0\\ \Leftrightarrow 2\left( {{{\cos }^3}2x - 1} \right) + \cos 2x\left( {\cos 2x - 1} \right) = 0\\ \Leftrightarrow 2\left( {\cos 2x - 1} \right)\left( {{{\cos }^2}2x + \cos 2x + 1} \right) + \cos 2x\left( {\cos 2x - 1} \right) = 0\\ \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {2{{\cos }^2}2x + 2\cos 2x + 2 + \cos 2x} \right) = 0\\ \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {2{{\cos }^2}2x + 3\cos 2x + 2} \right) = 0\\ \Leftrightarrow \cos 2x = 1 \Leftrightarrow 2x = k2\pi \Leftrightarrow x = k\pi \,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy nghiệm của phương trình đã cho là: \(x = k\pi \,\,\left( {k \in \mathbb{Z}} \right)\).
Giải phương trình \(4\sin x\sin \left( {x + \dfrac{\pi }{3}} \right)\sin \left( {x + \dfrac{{2\pi }}{3}} \right) + \cos 3x = 1\).
\(\begin{array}{l}
4\sin x\sin \left( {x + \dfrac{\pi }{3}} \right)\sin \left( {x + \dfrac{{2\pi }}{3}} \right) + \cos 3x = 1\\
\Leftrightarrow 4\sin x.\left( { - \dfrac{1}{2}} \right)\left[ {\cos \left( {2x + \pi } \right) - \cos \left( { - \dfrac{\pi }{3}} \right)} \right] + \cos 3x = 1\\
\Leftrightarrow - 2\sin x\left( { - \cos 2x - \dfrac{1}{2}} \right) + \cos 3x = 1\\
\Leftrightarrow 2\sin x\cos 2x + \sin x + \cos 3x = 1\\
\Leftrightarrow \sin 3x - \sin x + \sin x + \cos 3x = 1\\
\Leftrightarrow \sin 3x + \cos 3x = 1\\
\Leftrightarrow \dfrac{1}{{\sqrt 2 }}\sin 3x + \dfrac{1}{{\sqrt 2 }}\cos 3x = \dfrac{1}{{\sqrt 2 }}\\
\Leftrightarrow \sin \left( {3x + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
3x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}\)
Vậy phương trình có nghiệm là:
\(\left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in \mathbb{Z} } \right)\)
Số nghiệm của phương trình $\sin 5 x+\sqrt{3} \cos 5 x=2 \sin 7 x$ trên khoảng $\left(0 ; \dfrac{\pi}{2}\right)$ là
4
4
4
Phương trình
$\Leftrightarrow \dfrac{1}{2} \sin 5 x+\dfrac{\sqrt{3}}{2} \cos 5 x=\sin 7 x\\ \Leftrightarrow \sin \left(5 x+\dfrac{\pi}{3}\right)=\sin 7 x$
$\Leftrightarrow \sin 7 x=\sin \left(5 x+\dfrac{\pi}{3}\right) \Leftrightarrow\left[\begin{array}{l}7 x=5 x+\dfrac{\pi}{3}+k 2 \pi \\ 7 x=\pi-\left(5 x+\dfrac{\pi}{3}\right)+k 2 \pi\end{array} \Leftrightarrow\left[\begin{array}{l}x=\dfrac{\pi}{6}+k \pi \\ x=\dfrac{\pi}{18}+\dfrac{k \pi}{6}\end{array}(k \in \mathbb{Z})\right.\right.$
Xét khoảng $\left(0 ; \dfrac{\pi}{2}\right)$ :
+) $0<\dfrac{\pi}{6}+k \pi<\dfrac{\pi}{2} \Leftrightarrow-\dfrac{1}{6}<k<\dfrac{1}{3} \stackrel{k \in \mathbb{Z}}{\longrightarrow} k=0 \Rightarrow x=\dfrac{\pi}{6}$
+) $0<\dfrac{\pi}{18}+k \dfrac{\pi}{6}<\dfrac{\pi}{2} \Leftrightarrow-\dfrac{1}{3}<k<\dfrac{8}{3} \stackrel{k \in \mathbb{Z}}{\longrightarrow}\left[\begin{array}{l}k=0 \Rightarrow x=\dfrac{\pi}{18} \\ k=1 \Rightarrow x=\dfrac{2 \pi}{9} . \\ k=2 \Rightarrow x=\dfrac{7 \pi}{18}\end{array}\right.$
Vậy phương trình có 4 nghiệm thỏa mãn yêu cầu bài toán
Giải phương trình \(\cos 3x\tan 5x = \sin 7x\).
ĐKXĐ: \(\cos 5x \ne 0 \Leftrightarrow 5x \ne \dfrac{\pi }{2} + m\pi \Leftrightarrow x \ne \dfrac{\pi }{{10}} + \dfrac{{m\pi }}{5}\,\,\left( {m \in \mathbb{Z}} \right)\).
\(\begin{array}{l}\,\,\,\,\,\,\cos 3x\tan 5x = \sin 7x\\ \Leftrightarrow \cos 3x\sin 5x = \sin 7x\cos 5x\\ \Leftrightarrow \dfrac{1}{2}\left( {\sin 8x + \sin 2x} \right) = \dfrac{1}{2}\left( {\sin 12x + \sin 2x} \right)\\ \Leftrightarrow \sin 8x + \sin 2x = \sin 12x + \sin 2x\\ \Leftrightarrow \sin 12x = \sin 8x\\ \Leftrightarrow \left[ \begin{array}{l}12x = 8x + k2\pi \\12x = \pi - 8x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x = \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{{10}}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Đối chiếu điều kiện ta có:
\(\begin{array}{l}\,\,\,\,\,\,\dfrac{{k\pi }}{2} \ne \dfrac{\pi }{{10}} + \dfrac{{m\pi }}{5}\,\,\left( {k,\,\,m \in \mathbb{Z}} \right)\\ \Leftrightarrow 5k \ne 1 + 2m\\ \Leftrightarrow k \ne \dfrac{{1 + 2m}}{5}\end{array}\)
Do \(k \in \mathbb{Z}\) nên: \(k = \dfrac{{1 + 2m}}{5} \Leftrightarrow \dfrac{{1 + 2m}}{5}\) là số nguyên. Mà \(1 + 2m\) luôn lẻ nên \(\dfrac{{1 + 2m}}{5}\) không chia hết cho 2 với mọi \(m\). Do đó, nếu \(k \ne \dfrac{{1 + 2m}}{5}\) thì \(k\) phải là số nguyên chẵn.
\( \Rightarrow k\) chẵn, đặt \(k = 2n\), khi đó ta có \(x = \dfrac{{2n\pi }}{2} = n\pi \) \(\left( {n \in \mathbb{Z}} \right)\).
\(\begin{array}{l}\,\,\,\,\,\,\dfrac{\pi }{{20}} + \dfrac{{k\pi }}{{10}} \ne \dfrac{\pi }{{10}} + \dfrac{{m\pi }}{5}\,\,\left( {k,\,\,m \in \mathbb{Z}} \right)\\ \Leftrightarrow 1 + 2k \ne 2 + 4m\end{array}\)
Vì \(1 + 2k\) lẻ, \(2 + 4m\) chẵn nên \(1 + 2k \ne 2 + 4m\) luôn đúng với mọi \(k,\,\,m \in \mathbb{Z}\).
Vậy nghiệm của phương trình đã cho là: \(x = n\pi ;\,\,x = \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{{10}}\,\,\left( {k,\,\,n \in \mathbb{Z}} \right)\).
Giải phương trình \(8\sin x = \dfrac{{\sqrt 3 }}{{\cos x}} + \dfrac{1}{{\sin x}}\).
ĐKXĐ: \(\left\{ \begin{array}{l}\sin x \ne 0\\\cos x \ne 0\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\).
\(\begin{array}{l}\,\,\,\,\,\,8\sin x = \dfrac{{\sqrt 3 }}{{\cos x}} + \dfrac{1}{{\sin x}}\\ \Leftrightarrow 8{\sin ^2}x\cos x = \sqrt 3 \sin x + \cos x\\ \Leftrightarrow 4\sin x\sin 2x = \sqrt 3 \sin x + \cos x\\ \Leftrightarrow - 2\left( {\cos 3x - \cos x} \right) = \sqrt 3 \sin x + \cos x\\ \Leftrightarrow - 2\cos 3x + 2\cos x = \sqrt 3 \sin x + \cos x\\ \Leftrightarrow \cos x - \sqrt 3 \sin x = 2\cos 3x\\ \Leftrightarrow \dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = \cos 3x\\ \Leftrightarrow \cos x\cos \dfrac{\pi }{3} - \sin x\sin \dfrac{\pi }{3} = \cos 3x\\ \Leftrightarrow \cos \left( {x + \dfrac{\pi }{3}} \right) = \cos 3x\\ \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{3} = 3x + k2\pi \\x + \dfrac{\pi }{3} = - 3x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k\pi \\x = - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\,\,\left( {tm} \right)\end{array}\)
Vậy nghiệm của phương trình đã cho là: \(x = \dfrac{\pi }{6} + k\pi ;\,\,x = - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\,\,\left( {k \in \mathbb{Z}} \right)\).
Giải phương trình \(\sin 3x - \dfrac{2}{{\sqrt 3 }}{\sin ^2}x = 2\sin x\cos 2x\).
\(\begin{array}{l}\,\,\,\,\,\sin 3x - \dfrac{2}{{\sqrt 3 }}{\sin ^2}x = 2\sin x\cos 2x\\ \Leftrightarrow \sqrt 3 \sin 3x - 2{\sin ^2}x = \sqrt 3 \left( {\sin 3x - \sin x} \right)\\ \Leftrightarrow \sqrt 3 \sin 3x - 2{\sin ^2}x = \sqrt 3 \sin 3x - \sqrt 3 \sin x\\ \Leftrightarrow 2{\sin ^2}x - \sqrt 3 \sin x = 0\\ \Leftrightarrow \sin x\left( {2\sin x - \sqrt 3 } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\sin x = \dfrac{{\sqrt 3 }}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \dfrac{\pi }{3} + k2\pi \\x = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy nghiệm của phương trình đã cho là: \(x = k\pi ;\,\,x = \dfrac{\pi }{3} + k2\pi ;\,\,x = \dfrac{{2\pi }}{3} + k2\pi \,\,\left( {k \in \mathbb{Z}} \right)\).
Giải phương trình \(\left( {\sin x + \sqrt 3 \cos x} \right).\sin 3x = 2\).
\(\begin{array}{l}\,\,\,\,\,\left( {\sin x + \sqrt 3 \cos x} \right).\sin 3x = 2\\ \Leftrightarrow \left( {\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x} \right).\sin 3x = 1\\ \Leftrightarrow \left( {\sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3}} \right)\sin 3x = 1\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right)\sin 3x = 1\\ \Leftrightarrow - \dfrac{1}{2}\left[ {\cos \left( {4x + \dfrac{\pi }{3}} \right) - \cos \left( {2x - \dfrac{\pi }{3}} \right)} \right] = 1\\ \Leftrightarrow \cos \left( {4x + \dfrac{\pi }{3}} \right) - \cos \left( {2x - \dfrac{\pi }{3}} \right) =- 2\end{array}\)
Do \( - 1 \le \cos \left( {4x + \dfrac{\pi }{3}} \right),\,\,\cos \left( {2x - \dfrac{\pi }{3}} \right) \le 1\) nên
\(\begin{array}{l}\left\{ \begin{array}{l}\cos \left( {4x + \dfrac{\pi }{3}} \right) = -1\\\cos \left( {2x - \dfrac{\pi }{3}} \right) = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}4x + \dfrac{\pi }{3} = \pi +k2\pi \\2x - \dfrac{\pi }{3} = k2\pi \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}4x = \dfrac{2\pi }{3} + k2\pi \\2x = \dfrac{{\pi }}{3} + k2\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{\pi }{{6}} + \dfrac{{k\pi }}{2}\\x = \dfrac{{\pi }}{6} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right) \\\Rightarrow x =\dfrac{{\pi }}{6} + k\pi \end{array}\)
Vậy phương trình đã cho có họ nghiệm là \(x =\dfrac{{\pi }}{6} + k\pi\)
Giải phương trình \(\sin 18x\cos 13x = \sin 9x\cos 4x\).
\(\begin{array}{l}\,\,\,\,\,\sin 18x\cos 13x = \sin 9x\cos 4x\\ \Leftrightarrow \dfrac{1}{2}\left( {\sin 31x + \sin 5x} \right) = \dfrac{1}{2}\left( {\sin 13x + \sin 5x} \right)\\ \Leftrightarrow \sin 31x + \sin 5x = \sin 13x + \sin 5x\\ \Leftrightarrow \sin 31x = \sin 13x\\ \Leftrightarrow \left[ \begin{array}{l}31x = 13x + k2\pi \\31x = \pi - 13x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}18x = k2\pi \\44x = \pi + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{k\pi }}{9}\\x = \dfrac{\pi }{{44}} + \dfrac{{k\pi }}{{22}}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy nghiệm của phương trình là \(x = \dfrac{{k\pi }}{9};\,\,x = \dfrac{\pi }{{44}} + \dfrac{{k\pi }}{{22}}\,\,\left( {k \in \mathbb{Z}} \right)\)
Giải phương trình \(1 + \sin x + \cos 3x = \cos x + \sin 2x + \cos 2x\).
\(\begin{array}{l}\,\,\,\,\,1 + \sin x + \cos 3x = \cos x + \sin 2x + \cos 2x\\ \Leftrightarrow \left( {1 - \cos 2x} \right) + \left( {\sin x - \sin 2x} \right) + \left( {\cos 3x - \cos x} \right) = 0\\ \Leftrightarrow 2{\sin ^2}x + \left( {\sin x - \sin 2x} \right) - 2\sin 2x\sin x = 0\\ \Leftrightarrow 2\sin x\left( {\sin x - \sin 2x} \right) + \left( {\sin x - \sin 2x} \right) = 0\\ \Leftrightarrow \left( {\sin x - \sin 2x} \right)\left( {2\sin x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = \sin x\\\sin x =- \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x = x + k2\pi \\2x = \pi - x + k2\pi \\x = -\dfrac{\pi }{6} + k2\pi \\x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\x =- \dfrac{\pi }{{6}} + k2\pi \\x = \dfrac{{7\pi }}{{6}} + k2\pi \end{array} \right.\end{array}\)
Vậy nghiệm của phương trình là: \(x = k2\pi \), \(x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\), \(x = -\dfrac{\pi }{{6}} + k2\pi \), \(x = \dfrac{{7\pi }}{{6}} + k2\pi \).
Giải phương trình \(\cos x + \cos 3x + 2\cos 5x = 0\).
\(\begin{array}{l}\,\,\,\,\,\cos x + \cos 3x + 2\cos 5x = 0\\ \Leftrightarrow \cos x + \cos 3x + \cos 5x + \cos 5x = 0\\ \Leftrightarrow \left( {\cos x + \cos 5x} \right) + \left( {\cos 3x + \cos 5x} \right) = 0\\ \Leftrightarrow 2\cos 3x\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right)\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\cos x\left( {4{{\cos }^2}x - 3} \right)\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\cos x\left[ {\left( {4{{\cos }^2}x - 3} \right)\cos 2x + \cos 4x} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {\left[ {2\left( {1 + \cos 2x} \right) - 3} \right]\cos 2x + 2{{\cos }^2}2x - 1} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {\left( {2\cos 2x - 1} \right)\cos 2x + 2{{\cos }^2}2x - 1} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {4{{\cos }^2}2x - \cos 2x - 1} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\cos 2x = \dfrac{{1 + \sqrt {17} }}{8}\\\cos 2x = \dfrac{{1 - \sqrt {17} }}{8}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\2x = \pm \arccos \dfrac{{1 + \sqrt {17} }}{8} + k2\pi \\2x = \pm \arccos \dfrac{{1 - \sqrt {17} }}{8} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x = \pm \dfrac{1}{2}\arccos \dfrac{{1 + \sqrt {17} }}{8} + k\pi \\x = \pm \dfrac{1}{2}\arccos \dfrac{{1 - \sqrt {17} }}{8} + k\pi \end{array} \right.\end{array}\)
Vậy nghiệm của phương trình là: \(x = \dfrac{\pi }{2} + k\pi \), \(x = \pm \dfrac{1}{2}\arccos \dfrac{{1 + \sqrt {17} }}{8} + k\pi \), \(x = \pm \dfrac{1}{2}\arccos \dfrac{{1 - \sqrt {17} }}{8} + k\pi \).
Giải phương trình \(\sin 3x - \sin x + \sin 2x = 0\).
\(\begin{array}{l}\,\,\,\,\,\,\sin 3x - \sin x + \sin 2x = 0\\ \Leftrightarrow 2\cos 2x\sin x + 2\sin x\cos x = 0\\ \Leftrightarrow 2\sin x\left( {\cos 2x + \cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\cos 2x = - \cos x = \cos \left( {\pi - x} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\2x = \pi - x + k2\pi \\2x = x - \pi + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\x = - \pi + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy nghiệm của phương trình là: \(x = k\pi \), \(x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\).
Gọi m, M lần lượt là GTNN và GTLN của hàm số \(y = \dfrac{{\sin x + 3}}{{\sin x + \cos x + 2}}\). Khi đó giá trị của biểu thức m+M bằng
Ta có:
\(\begin{array}{l}y = \dfrac{{\sin x + 3}}{{\sin x + \cos x + 2}}\\ \Leftrightarrow y\left( {\sin x + \cos x + 2} \right) = \sin x + 3\\ \Leftrightarrow y.\cos x + \left( {y - 1} \right).\sin x = 3 - 2y\end{array}\)
Phương trình trên có nghiệm
\(\begin{array}{l} \Leftrightarrow {y^2} + {\left( {y - 1} \right)^2} \ge {\left( {3 - 2y} \right)^2}\\ \Leftrightarrow 2{y^2} - 2y + 1 \ge 9 - 12y + 4{y^2}\\ \Leftrightarrow 2{y^2} - 10y + 8 \le 0\\ \Leftrightarrow 1 \le y \le 4\end{array}\)
=> Min y=1, Max y=4