d, \(cot\left(\dfrac{\pi}{4}+x\right)=1\)
\(\Leftrightarrow cot\left(\dfrac{\pi}{4}+x\right)=cot\dfrac{\pi}{4}\)
\(\Leftrightarrow\dfrac{\pi}{4}+x=\dfrac{\pi}{4}+k\pi\)
\(\Leftrightarrow x=k\pi\)
e, \(sin\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^2-4x=k\pi\)
\(\Leftrightarrow\left(x-2\right)^2=k\pi+4\)
TH1: \(k\pi+4< 0\Leftrightarrow k< -\dfrac{4}{\pi}\Rightarrow\) phương trình vô nghiệm.
TH2: \(k\pi+4\ge0\Leftrightarrow k\ge-\dfrac{4}{\pi}\Rightarrow k\ge-1\)
Phương trình tương đương:
\(x-2=\pm\sqrt{k\pi+4}\)
\(\Leftrightarrow x=2\pm\sqrt{k\pi+4}\left(k\ge-1;k\in Z\right)\)
f, \(sin\left(x^2-x\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=x+\dfrac{\pi}{3}+k2\pi\\x^2-x=\pi-x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=\dfrac{\pi}{3}+k2\pi\\x^2=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=\dfrac{\pi}{3}+1+k2\pi\\x^2=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
TH1: \(\left(x-1\right)^2=\dfrac{\pi}{3}+1+k2\pi\)
Nếu \(\dfrac{\pi}{3}+1+k2\pi< 0\), phương trình vô nghiệm.
Nếu \(\dfrac{\pi}{3}+1+k2\pi\ge0\)
Phương trình tương đương:
\(x=1\pm\sqrt{\dfrac{\pi}{3}+1+k2\pi}\left(k\in N\right)\)
TH2: \(x^2=\dfrac{2\pi}{3}+k2\pi\)
Nếu \(\dfrac{2\pi}{3}+k2\pi< 0\), phương trình vô nghiệm.
Nếu \(\dfrac{2\pi}{3}+k2\pi\ge0\)
Phương trình tương đương:
\(x=\pm\sqrt{\dfrac{2\pi}{3}+k2\pi}\left(k\in N\right)\)
g, \(tan\left(x^2+2x+3\right)=tan2\)
\(\Leftrightarrow x^2+2x+3=2+k\pi\)
\(\Leftrightarrow\left(x+1\right)^2=k\pi\)
Nếu \(k\pi< 0\), phương trình vô nghiệm
Nếu \(k\pi\ge0\Leftrightarrow k\ge0\)
Phương trình tương đương:
\(\Leftrightarrow x+1=\pm k\pi\)
\(\Leftrightarrow x=-1\pm k\pi\left(k\in N\right)\)
h, \(cos\left(x^2+1\right)=0\)
\(\Leftrightarrow x^2+1=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x^2=\dfrac{\pi}{2}-1+k\pi\)
Làm tiếp tương tự các bài trên.
i, \(cos\left(x^2+x\right)=cos\left(x-1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=x-1+k2\pi\\x^2+x=1-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1+k2\pi\\\left(x+1\right)^2=2+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)