Giải phương trình: \(1-5\sin x+2\cos^2x=0\).
\(x=\dfrac{\pi}{6}+k2\pi,\ x=\dfrac{5\pi}{6}+k2\pi\).\(x=\dfrac{\pi}{6}+k\pi,\ x=\dfrac{5\pi}{6}+k\pi\).\(x=\dfrac{\pi}{3}+k2\pi,\ x=\dfrac{2\pi}{3}+k2\pi\).\(x=\dfrac{\pi}{3}+k\pi,\ x=\dfrac{2\pi}{3}+k\pi\).Hướng dẫn giải:\(pt \Leftrightarrow 1-5\sin x+2(1-\sin^2x)=0\)
\(\Leftrightarrow 2\sin^2x+5\sin x-3=0\)
\(\Leftrightarrow \left[ \begin{array}{} \sin x=-3(vô nghiệm)\\ \sin x=\dfrac{1}{2}=\sin\dfrac{\pi}{6} \end{array}{} \right.\)
\(\Leftrightarrow \left[ \begin{array}{} x=\dfrac{\pi}{6}+k2\pi\\ x=\dfrac{5\pi}{6}+k2\pi \end{array}{} \right.\)