Giải bài 1.14 trang 23 - SBT Đại số và Giải tích lớp 11
Giải các phương trình:
\(a)\sin 3x=-\dfrac{\sqrt{3}}{2}\)
\(b)\sin \left( 2x-{{15}^{o}} \right)=\dfrac{\sqrt{2}}{2}\)
\( c)\sin \left( \dfrac{x}{2}+{{10}^{o}} \right)=-\dfrac{1}{2}\)
\(d)\sin 4x=\dfrac{2}{3}\)
\(a)\sin 3x=-\dfrac{\sqrt{3}}{2} \\ \Leftrightarrow \left[ \begin{aligned} & 3x=-\dfrac{\pi }{3}+k2\pi \\ & 3x=\pi -\left( -\dfrac{\pi }{3} \right)+k2\pi \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x=-\dfrac{\pi }{9}+\dfrac{k2\pi }{3} \\ & x=\dfrac{4\pi }{9}+\dfrac{k2\pi }{3} \\ \end{aligned} \right.\,\,(k\in \mathbb{Z})\)
\(b)\sin \left( 2x-{{15}^{o}} \right)=\dfrac{\sqrt{2}}{2} \\ \Leftrightarrow \sin \left( 2x-{{15}^{o}} \right)=\sin {{45}^{o}} \\ \Leftrightarrow \left[ \begin{aligned} & 2x-{{15}^{o}}={{45}^{o}}+k{{.360}^{o}} \\ & 2x-{{15}^{o}}={{180}^{o}}-{{45}^{o}}+k{{.360}^{o}} \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x={{30}^{o}}+k{{.180}^{o}} \\ & x={{75}^{o}}+k{{.180}^{o}} \\ \end{aligned} \right.,(k\in \mathbb{Z})\)
\(c)\sin \left( \dfrac{x}{2}+{{10}^{o}} \right)=-\dfrac{1}{2} \\ \Leftrightarrow \sin \left( \dfrac{x}{2}+{{10}^{o}} \right)=\sin \left( -{{30}^{o}} \right) \\ \Leftrightarrow \left[ \begin{aligned} & \dfrac{x}{2}+{{10}^{o}}=-{{30}^{o}}+k{{.360}^{o}} \\ & \dfrac{x}{2}+{{10}^{o}}={{180}^{o}}-\left( -{{30}^{o}} \right)+k{{.360}^{o}} \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x=-{{80}^{o}}+k{{.720}^{o}} \\ & x={{400}^{o}}+k{{.720}^{o}} \\ \end{aligned} \right.,\,\,\left( k\in \mathbb{Z} \right)\)
\(d)\sin 4x=\dfrac{2}{3} \\ \Leftrightarrow \left[ \begin{aligned} & 4x=\arcsin \dfrac{2}{3}+k2\pi \\ & 4x=\pi -\arcsin \dfrac{2}{3}+k2\pi \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x=\dfrac{1}{4}\arcsin \dfrac{2}{3}+\dfrac{k\pi }{2} \\ & x=\dfrac{\pi }{4}-\dfrac{1}{4}\arcsin \dfrac{2}{3}+\dfrac{k\pi }{2} \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \)
Ghi nhớ:
\(\sin x=m \\ \Leftrightarrow \left[ \begin{aligned} & x=\alpha+k2\pi \\ & x=\pi -\alpha+k2\pi \\ \end{aligned} \right.\) với \(\sin\alpha =m\)
\(\sin f(x)=\sin g(x) \\ \Leftrightarrow \left[ \begin{aligned} & f(x)=g(x)+k2\pi \\ & f(x)=\pi -g(x)+k2\pi \\ \end{aligned} \right.\) (\(k\in \mathbb Z\))