12.
ĐK: \(x\ne\dfrac{k\pi}{2}\)
\(cosx+sinx+\dfrac{1}{cosx}+\dfrac{1}{sinx}+tanx+cotx=-2\)
\(\Leftrightarrow cosx+sinx+\dfrac{cosx+sinx}{cosx.sinx}+\dfrac{sin^2x+cos^2x+2sinx.cosx}{sinx.cosx}=0\)
\(\Leftrightarrow cosx+sinx+\dfrac{cosx+sinx}{cosx.sinx}+\dfrac{\left(sinx+cosx\right)^2}{sinx.cosx}=0\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(1+\dfrac{1}{cosx.sinx}+\dfrac{sinx+cosx}{sinx.cosx}\right)=0\)
\(\Leftrightarrow\dfrac{\left(cosx+sinx\right)\left(sinx+1\right)\left(cosx+1\right)}{sinx.cosx}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx+sinx=0\\sinx+1=0\\cosx+1=0\end{matrix}\right.\)
TH1: \(cosx+sinx=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\left(tm\right)\)
TH2: \(sinx+1=0\)
\(\Leftrightarrow sinx=-1\)
\(\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\left(l\right)\)
TH3: \(cosx+1=0\)
\(\Leftrightarrow cosx=-1\)
\(\Leftrightarrow x=\pi+k2\pi\left(l\right)\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{4}+k\pi\)
14.
\(cos^2\left(\dfrac{\pi}{3}+x\right)+4cos\left(\dfrac{\pi}{6}-x\right)=4\)
\(\Leftrightarrow sin^2\left(\dfrac{\pi}{6}-x\right)+4cos\left(\dfrac{\pi}{6}-x\right)-4=0\)
\(\Leftrightarrow cos^2\left(\dfrac{\pi}{6}-x\right)+4cos\left(\dfrac{\pi}{6}-x\right)-5=0\)
\(\Leftrightarrow\left[cos\left(\dfrac{\pi}{6}-x\right)-1\right].\left[cos\left(\dfrac{\pi}{6}-x\right)+5\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{\pi}{6}-x\right)=1\\cos\left(\dfrac{\pi}{6}-x\right)=-5\left(l\right)\end{matrix}\right.\)
\(cos\left(\dfrac{\pi}{6}-x\right)=1\Leftrightarrow\dfrac{\pi}{6}-x=0\Leftrightarrow x=\dfrac{\pi}{6}\left(tm\right)\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{6}\)
