Câu 1:
\(cos7x-\sqrt{3}sin7x=-2\\ \Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=-1\\ \Leftrightarrow7x+\dfrac{\pi}{3}=-\pi+k2\pi\\ \Leftrightarrow x=-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\)
Vì \(x\in[\dfrac{2\pi}{5};\dfrac{6\pi}{7}]\)
\(\Rightarrow\dfrac{2\pi}{5}\le x\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{2\pi}{5}\le-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{31}{15}\le k\le\dfrac{11}{3}\)
Vì \(k\in Z\) nên \(k=3\)
Vậy \(x\) cần tìm là \(\dfrac{2\pi}{3}\)
Câu 2:
\(2sin^2x-sinxcosx-cos^2x=m\\ \Leftrightarrow2\dfrac{1-cos2x}{2}-\dfrac{1}{2}s\text{in2}x-\dfrac{1+cos2x}{2}=m\\ \Leftrightarrow3cos2x+s\text{in2}x=1-2m\)
Điều kiện để phương trình có nghiệm là:
\(3^2+1^2\ge\left(1-2m\right)^2\\ \Leftrightarrow4m^2-4m-9\le0\\ \Leftrightarrow\dfrac{1-\sqrt{10}}{2}\le m\le\dfrac{1+\sqrt{10}}{2}\)
