a/ ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\le\frac{3}{4}\\sinx\ge0\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{3-4cos2x}=\sqrt{2sinx}\)
\(\Leftrightarrow3-4cos2x=2sinx\)
\(\Leftrightarrow3-4\left(1-2sin^2x\right)=2sinx\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{4}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\sqrt{4sin^2x-1}=3sinx-1\) (\(\frac{1}{3}\le sinx\le1\))
\(\Leftrightarrow4sin^2x-1=9sin^2x-6sinx+1\)
\(\Leftrightarrow5sin^2x-6sinx+2=0\)
Phương trình vô nghiệm
Đúng 0
Bình luận (0)
