a) 2cos^2 2x +cos2 x =2sin^2x +1/2 b) 2sin^2 3x +2 =(cos2x+1)(4sin2x-1) 11/09/2021 Bởi Sadie a) 2cos^2 2x +cos2 x =2sin^2x +1/2 b) 2sin^2 3x +2 =(cos2x+1)(4sin2x-1)
\(\begin{array}{l} a)2{\cos ^2}2x + \cos 2x = 2{\sin ^2}x + \frac{1}{2}\\ \Rightarrow 2{\cos ^2}2x + \cos 2x = 1 – \cos 2x + \frac{1}{2}\\ \Rightarrow 2{\cos ^2}2x + 2\cos 2x – \frac{3}{2} = 0\\ \Rightarrow \left[ \begin{array}{l} \cos x = \frac{1}{2}\\ \cos x = \frac{{ – 3}}{2} < - 1(l) \end{array} \right.\\ \Rightarrow x = \pm \frac{\pi }{3} + k2\pi (k \in ) \end{array}\) Bình luận
\(\begin{array}{l}
a)2{\cos ^2}2x + \cos 2x = 2{\sin ^2}x + \frac{1}{2}\\
\Rightarrow 2{\cos ^2}2x + \cos 2x = 1 – \cos 2x + \frac{1}{2}\\
\Rightarrow 2{\cos ^2}2x + 2\cos 2x – \frac{3}{2} = 0\\
\Rightarrow \left[ \begin{array}{l}
\cos x = \frac{1}{2}\\
\cos x = \frac{{ – 3}}{2} < - 1(l) \end{array} \right.\\ \Rightarrow x = \pm \frac{\pi }{3} + k2\pi (k \in ) \end{array}\)