Đặt \(x+\dfrac{\pi}{3}=t\Rightarrow\dfrac{\pi}{6}-x=\dfrac{\pi}{2}-t\)
Ta được:
\(cos2t+4cos\left(\dfrac{\pi}{2}-t\right)=\dfrac{5}{2}\)
\(\Leftrightarrow1-2sin^2t+4sint=\dfrac{5}{2}\)
\(\Leftrightarrow4sin^2t-8sint+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sint=\dfrac{3}{2}>1\left(ktm\right)\\sint=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
C48)
Tại \(x\in\left(-\dfrac{\pi}{2};0\right)\Rightarrow sinx\in\left(-1;0\right)\)
Đặt \(t=sinx\) \(;t\in\left(-1;0\right)\)
Pttt: \(2t^2-\left(2m+1\right)t+m=0\) (*)
Để pt ban đầu có nghiệm \(\in\left(-\dfrac{\pi}{2};0\right)\) khi pt (*) có nghiệm t thỏa mãn \(-1< t< 0\)
Có \(\Delta=\left(2m-1\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}t_1=\dfrac{2m+1-\left(2m-1\right)}{4}=\dfrac{1}{2}\\t_2=\dfrac{2m+1+\left(2m-1\right)}{4}=m\end{matrix}\right.\)
Vì \(t_1=\dfrac{1}{2}\notin\left(-1;0\right)\)
Để pt có nghiệm thỏa mãn đk\(\Leftrightarrow m\in\left(-1;0\right)\)
Ý A
