tìm cực trị y=sinx+sos2x toán 12 ạ hàm số

\[\begin{array}{l} y = \sin x + \cos 2x\\ \Rightarrow y' = \cos x - 2\sin 2x \Rightarrow y'' = - \sin x - 4\cos 2x\\ \Rightarrow y' = 0 \Leftrightarrow \cos x - 2\sin 2x = 0\\ \Leftrightarrow \cos x - 4\sin x\cos x = 0\\ \Leftrightarrow \cos x\left( {1 - 4\sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = 0\\ \sin x = \frac{1}{4} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k\pi \\ x = \arcsin \frac{1}{4} + m2\pi \\ x = \pi - \arcsin \frac{1}{4} + m2\pi \end{array} \right.\\ \Rightarrow y''\left( {\frac{\pi }{2} + k\pi } \right) = - \sin \left( {\frac{\pi }{2} + k\pi } \right) - 4\cos 2\left( {\frac{\pi }{2} + k\pi } \right)\\ = - \sin \left( {\frac{\pi }{2} + k\pi } \right) - 4\cos \left( {\pi + k2\pi } \right) = - \sin \left( {\frac{\pi }{2} + k\pi } \right) + 4 = \left\{ \begin{array}{l} 3\,\,\,khi\,\,\,k\,\,chan\\ 5\,\,\,khi\,\,\,k\,\,\,le \end{array} \right.\\ \Rightarrow y''\left( {\frac{\pi }{2} + k\pi } \right) > \,\,0\,\,\forall k \Rightarrow x = \frac{\pi }{2} + k\pi \,\,\,la\,\,\,diem\,\,cuc\,\,tieu.\\ y''\left( {\arcsin \frac{1}{4} + m2\pi } \right) = - \sin \left( {\arcsin \frac{1}{4} + m2\pi } \right) - 4\cos 2\left( {\arcsin \frac{1}{4} + m2\pi } \right)\\ = - \frac{1}{4} - 4.\frac{7}{8} < 0\\ \Rightarrow x = \arcsin \frac{1}{4} + m2\pi \,\,\,\,la\,\,\,diem\,\,cuc\,\,dai.\\ y''\left( {\pi - \arcsin \frac{1}{4} + m2\pi } \right) = - \sin \left( {\pi - \arcsin \frac{1}{4} + m2\pi } \right) - 4\cos 2\left( {\pi - \arcsin \frac{1}{4} + m2\pi } \right)\\ = - \frac{1}{4} - \frac{7}{2} < 0\\ \Rightarrow x = \pi - \arcsin \frac{1}{4} + m2\pi \,\,\,\,la\,\,\,diem\,\,cuc\,\,dai. \end{array}\]