Giải bài 1.16 trang 24 - SBT Đại số và Giải tích lớp 11

$\begin{align} & a)\tan \left( 2x+{{45}^{o}} \right)=-1 \\ & \Leftrightarrow \tan \left( 2x+{{45}^{o}} \right)=\tan \left( -{{45}^{o}} \right) \\ & \Rightarrow 2x+{{45}^{o}}=-{{45}^{o}}+k{{.180}^{o}} \\ & \Leftrightarrow x=-{{45}^{o}}+k{{.90}^{o}}\,\,\left( k\in \mathbb{Z} \right) \\ \end{align}$

$\begin{align} & b)\cot \left( x+\dfrac{\pi }{3} \right)=\sqrt{3} \\ & \Leftrightarrow \cot \left( x+\dfrac{\pi }{3} \right)=\cot \dfrac{\pi }{6} \\ & \Rightarrow x+\dfrac{\pi }{3}=\dfrac{\pi }{6}+k\pi \\ & \Leftrightarrow x=\dfrac{-\pi }{6}+k\pi \,\,\left( k\in \mathbb{Z} \right) \\ \end{align}$

$\begin{align} & c)\tan \left( \dfrac{x}{2}-\dfrac{\pi }{4} \right)=\tan \dfrac{\pi }{8} \\ & \Rightarrow \dfrac{x}{2}-\dfrac{\pi }{4}=\dfrac{\pi }{8}+k\pi \\ & \Leftrightarrow x=\dfrac{3\pi }{4}+k2\pi \,\,\left( k\in \mathbb{Z} \right) \\ \end{align}$

$\begin{align} & d)\cot \left( \dfrac{x}{3}+{{20}^{o}} \right)=-\dfrac{\sqrt{3}}{3} \\ & \Leftrightarrow \cot \left( \dfrac{x}{3}+{{20}^{o}} \right)=\cot {{120}^{o}} \\ & \Rightarrow \dfrac{x}{3}+{{20}^{o}}={{120}^{o}}+k{{180}^{o}} \\ & \Leftrightarrow x={{300}^{o}}+k{{540}^{o}}\,\,\left( k\in \mathbb{Z} \right) \\ \end{align}$

Ghi nhớ:

$\tan x=\tan \alpha \Rightarrow x=\alpha+k\pi, \,\,k\in \mathbb Z$

$\cot x=\cot \alpha \Rightarrow x=\alpha+k\pi, \,\,k\in \mathbb Z$