04-12-2022 | 1641
Giải phương trình \(sin8x-cos6x=\sqrt{3}(sin6x+cos8x)\)
1. Hướng dẫn
Áp dụng công thức
\(sin(a \pm b) = sinacosb\pm cosasinb\)
\(cos(a\pm b) = cosacosb \mp sinasinb\)
Rút gọn biểu thức \(asinx + bcosx\) như sau:
Nhân cả tử và mẫu cho \(\sqrt{a^{2} + b^{2}}\), ta được:
\(asinx + bcosx =\) \(\sqrt{a^{2}+b^{2}}\left (\frac{a}{\sqrt{a^{2}+b^{2}}}sinx+ \frac{b}{\sqrt{a^{2}+b^{2}}}cosx \right )\)
Vì \(-1\leq \frac{a}{\sqrt{a^{2}+b^{2}}}\leq 1\), \(-1\leq \frac{b}{\sqrt{a^{2}+b^{2}}}\leq 1\) nên tồn tại số \(\varphi\) sao cho:
\(cos\varphi = \frac{a}{\sqrt{a^{2}+b^{2}}}\) \(\Rightarrow\) \(sin\varphi = \frac{b}{\sqrt{a^{2}+b^{2}}}\) vì \(\left (\frac{a}{\sqrt{a^{2}+b^{2}}} \right )^{2} + \left (\frac{b}{\sqrt{a^{2}+b^{2}}} \right )^{2}=1\)
\(asinx+bcosx = cos\varphi sinx + sin\varphi cosx\)
\(= sinx cos\varphi + cosxsin\varphi \)
\(= sin(x + \varphi )\)
2. Bài giải
\(\sin 8x + \cos 6x = \sqrt 3 (\sin 6x + \cos 8x) \)
\(\Leftrightarrow \sin 8x - \cos 6x = \sqrt 3 \sin 6x + \sqrt 3 \cos 8x\)
\(\Leftrightarrow \sin 8x - \sqrt 3 \cos 8x = \sqrt 3 \sin 6x + \cos 6x\)
\(\Leftrightarrow 2\left( {\frac{1}{2}\sin 8x - \frac{{\sqrt 3 }}{2}\cos 8x} \right) = 2\left( {\frac{{\sqrt 3 }}{2}\sin 6x + \frac{1}{2}\cos 6x} \right) \)
\(\Leftrightarrow \cos \frac{\pi }{3}\sin 8x - \sin \frac{\pi }{3}\cos 8x = \cos \frac{\pi }{6}\sin 6x + \sin \frac{\pi }{6}\cos 6x \)
\(\Leftrightarrow \sin 8x\cos \frac{\pi }{3} - \cos 8x\sin \frac{\pi }{3} = \sin 6x\cos \frac{\pi }{6} + \cos 6x\sin \frac{\pi }{6}\)
\(\Leftrightarrow \sin \left( {8x - \frac{\pi }{3}} \right) = \sin \left( {6x + \frac{\pi }{6}} \right)\)
\(\Leftrightarrow \left\{ \begin{gathered} 8x - \frac{\pi }{3} = 6x + \frac{\pi }{6} + k2\pi \hfill \\ 8x - \frac{\pi }{3} = \pi - \left( {6x + \frac{\pi }{6}} \right) + k2\pi \hfill \\ \end{gathered} \right.;\,\,k \in \mathbb{Z} \)
\(\Leftrightarrow \left\{ \begin{gathered} 2x = \frac{\pi }{2} + k2\pi \hfill \\ 14x = \frac{{7\pi }}{6} + k2\pi \hfill \\ \end{gathered} \right.;\,\,k \in \mathbb{Z} \)
\(\Leftrightarrow \left\{ \begin{gathered} x = \frac{\pi }{4} + k\pi \hfill \\ x = \frac{\pi }{{12}} + k\pi \hfill \\ \end{gathered} \right.;\,\,k \in \mathbb{Z} \)
