Phương trình lượng giác thường gặp

\(\begin{array}{l}\,\,\,\,\,\sqrt 3 \cos 5x - 2\sin 3x\cos 2x - \sin x = 0\\ \Leftrightarrow \sqrt 3 \cos 5x - \left( {\sin 5x + \sin x} \right) - \sin x = 0\\ \Leftrightarrow \sqrt 3 \cos 5x - \sin 5x = 2\sin x\\ \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos 5x - \dfrac{1}{2}\sin 5x = \sin x\\ \Leftrightarrow \sin \dfrac{\pi }{3}\cos 5x - \cos \dfrac{\pi }{3}\sin 5x = \sin x\\ \Leftrightarrow \sin \left( {\dfrac{\pi }{3} - 5x} \right) = \sin x\\ \Leftrightarrow \left[ \begin{array}{l}\dfrac{\pi }{3} - 5x = x + k2\pi \\\dfrac{\pi }{3} - 5x = \pi  - x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\\x =  - \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình là \(x = \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3};\,\,x =  - \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\left( {k \in \mathbb{Z}} \right)\).

\(\begin{array}{l}\,\,\,\,\,\,\cos x\cos \dfrac{x}{2}\cos \dfrac{{3x}}{2} - \sin x\sin \dfrac{x}{2}\sin \dfrac{{3x}}{2} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{1}{2}\cos x\left( {\cos 2x + \cos x} \right) + \dfrac{1}{2}\sin x\left( {\cos 2x - \cos x} \right) = \dfrac{1}{2}\\ \Leftrightarrow \cos x\cos 2x + {\cos ^2}x + \sin x\cos 2x - \sin x\cos x = 1\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\cos x + {\cos ^2}x - 1 = 0\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\cos x - {\sin ^2}x = 0\\ \Leftrightarrow \cos 2x\left( {\sin x + \cos x} \right) - \sin x\left( {\sin x + \cos x} \right) = 0\\ \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {\cos 2x - \sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x + \cos x = 0\\\cos 2x - \sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sin x =  - \cos x\\1 - 2{\sin ^2}x - \sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\tan x =  - 1\\\sin x = \dfrac{1}{2}\\\sin x =  - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \dfrac{\pi }{4} + k\pi \\x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \\x =  - \dfrac{\pi }{2} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình đã cho là: \(x =  - \dfrac{\pi }{4} + k\pi ;\,\,x = \dfrac{\pi }{6} + k2\pi ;\)\(x = \dfrac{{5\pi }}{6} + k2\pi ;\,\,x =  - \dfrac{\pi }{2} + k2\pi \)\(\left( {k \in \mathbb{Z}} \right)\).

\(\begin{array}{l}\,\,\,\,\,\,\cos 2x + \cos 4x + \cos 6x = \cos x\cos 2x\cos 3x + 2\\ \Leftrightarrow 2\cos 4x\cos 2x + \cos 4x = \dfrac{1}{2}\cos 2x\left( {\cos 4x + \cos 2x} \right) + 2\\ \Leftrightarrow 2\cos 4x\cos 2x + \cos 4x = \dfrac{1}{2}\cos 2x\cos 4x + \dfrac{1}{2}{\cos ^2}2x + 2\\ \Leftrightarrow \dfrac{3}{2}\cos 4x\cos 2x + \cos 4x = \dfrac{1}{2}{\cos ^2}2x + 2\\ \Leftrightarrow 3\cos 4x\cos 2x + 2\cos 4x = {\cos ^2}2x + 4\\ \Leftrightarrow 3\left( {2{{\cos }^2}2x - 1} \right)\cos 2x + 2\left( {2{{\cos }^2}2x - 1} \right) = {\cos ^2}2x + 4\\ \Leftrightarrow 6{\cos ^3}2x - 3\cos 2x + 4{\cos ^2}2x - 2 = {\cos ^2}2x + 4\\ \Leftrightarrow 6{\cos ^3}2x + 3{\cos ^2}2x - 3\cos 2x - 6 = 0\\ \Leftrightarrow 2{\cos ^3}2x + {\cos ^2}2x - \cos 2x - 2 = 0\\ \Leftrightarrow 2\left( {{{\cos }^3}2x - 1} \right) + \cos 2x\left( {\cos 2x - 1} \right) = 0\\ \Leftrightarrow 2\left( {\cos 2x - 1} \right)\left( {{{\cos }^2}2x + \cos 2x + 1} \right) + \cos 2x\left( {\cos 2x - 1} \right) = 0\\ \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {2{{\cos }^2}2x + 2\cos 2x + 2 + \cos 2x} \right) = 0\\ \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {2{{\cos }^2}2x + 3\cos 2x + 2} \right) = 0\\ \Leftrightarrow \cos 2x = 1 \Leftrightarrow 2x = k2\pi  \Leftrightarrow x = k\pi \,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình đã cho là: \(x = k\pi \,\,\left( {k \in \mathbb{Z}} \right)\).

\(\begin{array}{l}
4\sin x\sin \left( {x + \dfrac{\pi }{3}} \right)\sin \left( {x + \dfrac{{2\pi }}{3}} \right) + \cos 3x = 1\\
\Leftrightarrow 4\sin x.\left( { - \dfrac{1}{2}} \right)\left[ {\cos \left( {2x + \pi } \right) - \cos \left( { - \dfrac{\pi }{3}} \right)} \right] + \cos 3x = 1\\
\Leftrightarrow - 2\sin x\left( { - \cos 2x - \dfrac{1}{2}} \right) + \cos 3x = 1\\
\Leftrightarrow 2\sin x\cos 2x + \sin x + \cos 3x = 1\\
\Leftrightarrow \sin 3x - \sin x + \sin x + \cos 3x = 1\\
\Leftrightarrow \sin 3x + \cos 3x = 1\\
\Leftrightarrow \dfrac{1}{{\sqrt 2 }}\sin 3x + \dfrac{1}{{\sqrt 2 }}\cos 3x = \dfrac{1}{{\sqrt 2 }}\\
\Leftrightarrow \sin \left( {3x + \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
3x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}\)

Vậy phương trình có nghiệm là:

\(\left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in \mathbb{Z} } \right)\)

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: \(x = n\pi ;\,\,x = \dfrac{\pi }{{20}} + \dfrac{{k\pi }}{{10}}\,\,\left( {k,\,\,n \in \mathbb{Z}} \right)\).

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: \(x =   \dfrac{\pi }{6} + k\pi ;\,\,x =  - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\,\,\left( {k \in \mathbb{Z}} \right)\).

\(\begin{array}{l}\,\,\,\,\,\sin 3x - \dfrac{2}{{\sqrt 3 }}{\sin ^2}x = 2\sin x\cos 2x\\ \Leftrightarrow \sqrt 3 \sin 3x - 2{\sin ^2}x = \sqrt 3 \left( {\sin 3x - \sin x} \right)\\ \Leftrightarrow \sqrt 3 \sin 3x - 2{\sin ^2}x = \sqrt 3 \sin 3x - \sqrt 3 \sin x\\ \Leftrightarrow 2{\sin ^2}x - \sqrt 3 \sin x = 0\\ \Leftrightarrow \sin x\left( {2\sin x - \sqrt 3 } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\sin x = \dfrac{{\sqrt 3 }}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  k\pi \\x = \dfrac{\pi }{3} + k2\pi \\x = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình đã cho là: \(x =  k\pi ;\,\,x = \dfrac{\pi }{3} + k2\pi ;\,\,x = \dfrac{{2\pi }}{3} + k2\pi \,\,\left( {k \in \mathbb{Z}} \right)\).

\(\begin{array}{l}\,\,\,\,\,\left( {\sin x + \sqrt 3 \cos x} \right).\sin 3x = 2\\ \Leftrightarrow \left( {\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x} \right).\sin 3x = 1\\ \Leftrightarrow \left( {\sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3}} \right)\sin 3x = 1\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right)\sin 3x = 1\\ \Leftrightarrow  - \dfrac{1}{2}\left[ {\cos \left( {4x + \dfrac{\pi }{3}} \right) - \cos \left( {2x - \dfrac{\pi }{3}} \right)} \right] =   1\\ \Leftrightarrow \cos \left( {4x + \dfrac{\pi }{3}} \right) - \cos \left( {2x - \dfrac{\pi }{3}} \right) =- 2\end{array}\)

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 \(x =\dfrac{{\pi }}{6} + k\pi\)

\(\begin{array}{l}\,\,\,\,\,\sin 18x\cos 13x = \sin 9x\cos 4x\\ \Leftrightarrow \dfrac{1}{2}\left( {\sin 31x + \sin 5x} \right) = \dfrac{1}{2}\left( {\sin 13x + \sin 5x} \right)\\ \Leftrightarrow \sin 31x + \sin 5x = \sin 13x + \sin 5x\\ \Leftrightarrow \sin 31x = \sin 13x\\ \Leftrightarrow \left[ \begin{array}{l}31x = 13x + k2\pi \\31x = \pi  - 13x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}18x = k2\pi \\44x = \pi  + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{k\pi }}{9}\\x = \dfrac{\pi }{{44}} + \dfrac{{k\pi }}{{22}}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình là \(x = \dfrac{{k\pi }}{9};\,\,x = \dfrac{\pi }{{44}} + \dfrac{{k\pi }}{{22}}\,\,\left( {k \in \mathbb{Z}} \right)\)

\(\begin{array}{l}\,\,\,\,\,1 + \sin x + \cos 3x = \cos x + \sin 2x + \cos 2x\\ \Leftrightarrow \left( {1 - \cos 2x} \right) + \left( {\sin x - \sin 2x} \right) + \left( {\cos 3x - \cos x} \right) = 0\\ \Leftrightarrow 2{\sin ^2}x + \left( {\sin x - \sin 2x} \right) - 2\sin 2x\sin x = 0\\ \Leftrightarrow 2\sin x\left( {\sin x - \sin 2x} \right) + \left( {\sin x - \sin 2x} \right) = 0\\ \Leftrightarrow \left( {\sin x - \sin 2x} \right)\left( {2\sin x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 2x = \sin x\\\sin x =- \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x = x + k2\pi \\2x = \pi  - x + k2\pi \\x = -\dfrac{\pi }{6} + k2\pi \\x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k2\pi \\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\x =- \dfrac{\pi }{{6}} + k2\pi \\x = \dfrac{{7\pi }}{{6}} + k2\pi \end{array} \right.\end{array}\)

Vậy nghiệm của phương trình là: \(x = k2\pi \), \(x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\), \(x = -\dfrac{\pi }{{6}} + k2\pi \), \(x = \dfrac{{7\pi }}{{6}} + k2\pi \).

 \(\begin{array}{l}\,\,\,\,\,\cos x + \cos 3x + 2\cos 5x = 0\\ \Leftrightarrow \cos x + \cos 3x + \cos 5x + \cos 5x = 0\\ \Leftrightarrow \left( {\cos x + \cos 5x} \right) + \left( {\cos 3x + \cos 5x} \right) = 0\\ \Leftrightarrow 2\cos 3x\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right)\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\cos x\left( {4{{\cos }^2}x - 3} \right)\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\cos x\left[ {\left( {4{{\cos }^2}x - 3} \right)\cos 2x + \cos 4x} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {\left[ {2\left( {1 + \cos 2x} \right) - 3} \right]\cos 2x + 2{{\cos }^2}2x - 1} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {\left( {2\cos 2x - 1} \right)\cos 2x + 2{{\cos }^2}2x - 1} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {4{{\cos }^2}2x - \cos 2x - 1} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\cos 2x = \dfrac{{1 + \sqrt {17} }}{8}\\\cos 2x = \dfrac{{1 - \sqrt {17} }}{8}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\2x =  \pm \arccos \dfrac{{1 + \sqrt {17} }}{8} + k2\pi \\2x =  \pm \arccos \dfrac{{1 - \sqrt {17} }}{8} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 + \sqrt {17} }}{8} + k\pi \\x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 - \sqrt {17} }}{8} + k\pi \end{array} \right.\end{array}\)

Vậy nghiệm của phương trình là: \(x = \dfrac{\pi }{2} + k\pi \), \(x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 + \sqrt {17} }}{8} + k\pi \), \(x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 - \sqrt {17} }}{8} + k\pi \).

\(\begin{array}{l}\,\,\,\,\,\,\sin 3x - \sin x + \sin 2x = 0\\ \Leftrightarrow 2\cos 2x\sin x + 2\sin x\cos x = 0\\ \Leftrightarrow 2\sin x\left( {\cos 2x + \cos x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\cos 2x =  - \cos x = \cos \left( {\pi  - x} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\2x = \pi  - x + k2\pi \\2x = x - \pi  + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\\x =  - \pi  + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy nghiệm của phương trình là: \(x = k\pi \), \(x = \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}\).