\(\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x+7\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=4\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+7cos\left(x-\frac{\pi}{3}\right)=4\)
Đặt \(x-\frac{\pi}{3}=t\Rightarrow x=\frac{\pi}{3}+t\)
\(\Rightarrow cos\left(2t+\frac{2\pi}{3}+\frac{\pi}{3}\right)+7cost-4=0\)
\(\Leftrightarrow cos\left(2t+\pi\right)+7cost-4=0\)
\(\Leftrightarrow-cos2t+7cost-4=0\)
\(\Leftrightarrow-2cos^2t+7cost-3=0\)
\(\Leftrightarrow...\)
Để \(\lim\limits_{x\rightarrow2}\frac{f\left(x\right)+1}{x-2}=a\) hữu hạn thì tử số \(f\left(x\right)+1\) phải có nghiệm \(x=2\) hay \(f\left(2\right)+1=0\)
Ta có:
\(\lim\limits_{x\rightarrow2}\frac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\frac{f\left(x\right)+2x+1-x^2}{\left(x-2\right)\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}\)
\(=\lim\limits_{x\rightarrow2}\left[\frac{f\left(x\right)+1}{\left(x-2\right)\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}\right]-\lim\limits_{x\rightarrow2}\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{a}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}-\lim\limits_{x\rightarrow2}\frac{x}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}\)
\(=\frac{a}{4\left(\sqrt{0+2.2}+2\right)}-\frac{2}{4\left(\sqrt{0+2.2}+2\right)}=\frac{a}{16}-\frac{1}{8}=\frac{a-2}{16}\)
