Giải bài 1.25 trang 37 - SBT Đại số và Giải tích lớp 11

$\begin{aligned} & a)\cos 2x-\sin x-1=0 \\ & \Leftrightarrow 1-2{{\sin }^{2}}x-\sin x-1=0 \\ & \Leftrightarrow 2{{\sin }^{2}}x+\sin x=0 \\ & \Leftrightarrow \left[ \begin{aligned} & \sin x=0 \\ & \sin x=\dfrac{-1}{2} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=k\pi \\ & x=-\dfrac{\pi }{6}+k2\pi \\ & x=\dfrac{7\pi }{6}+k2\pi \\ \end{aligned} \right.\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$

$\begin{aligned} & b)\cos x\cos 2x=1+\sin x\sin 2x \\ & \Leftrightarrow \cos x\cos 2x-\sin x\sin 2x=1 \\ & \Leftrightarrow \cos 3x=1 \\ & \Leftrightarrow 3x=k2\pi \\ & \Leftrightarrow x=k\dfrac{2\pi }{3}\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$

$\begin{aligned} & c)\,4\sin x\cos x\cos 2x=-1 \\ & \Leftrightarrow 2\sin 2x\cos 2x=-1 \\ & \Leftrightarrow \sin 4x=-1 \\ & \Leftrightarrow 4x=-\dfrac{\pi }{2}+k2\pi \\ & \Leftrightarrow x=-\dfrac{\pi }{8}+\dfrac{k\pi }{2}\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$

$\begin{aligned} & d)\,\text{ĐKXĐ}:x\ne \dfrac{k\pi }{2} \\ & \tan x=3\cot x \\ & \Leftrightarrow \tan x=\dfrac{3}{\tan x} \\ & \Rightarrow {{\tan }^{2}}x=3 \\ & \Leftrightarrow \left[ \begin{aligned} & \tan x=\sqrt{3} \\ & \tan x=-\sqrt{3} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=\dfrac{\pi }{3}+k\pi \\ & x=-\dfrac{\pi }{3}+k\pi \\ \end{aligned} \right.\,\,\,\left( k\in \mathbb{Z} \right) \\ \end{aligned}$