Đáp án: A Giải thích các bước giải: \(\begin{array}{l}\cos \left( {\dfrac{\pi }{6} – x} \right) = \sin \left( {\dfrac{\pi }{2} – \dfrac{\pi }{6} + x} \right)\\ \to \cos \left( {\dfrac{\pi }{6} – x} \right) = \sin \left( {\dfrac{\pi }{3} + x} \right)\\Có:\cos 2\left( {x + \dfrac{\pi }{3}} \right) = 1 – 2{\sin ^2}\left( {x + \dfrac{\pi }{3}} \right)\\Pt \to 1 – 2{\sin ^2}\left( {x + \dfrac{\pi }{3}} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = \dfrac{5}{2}\\ \to 1 – 2.{\cos ^2}\left( {\dfrac{\pi }{6} – x} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = \dfrac{5}{2}\\ \to – 2.{\cos ^2}\left( {\dfrac{\pi }{6} – x} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) – \dfrac{3}{2} = 0\\Đặt:\cos \left( {\dfrac{\pi }{6} – x} \right) = t\\Pt \to – 2{t^2} + 4t – \dfrac{3}{2} = 0\\ \to – 4{t^2} + 8t – 3 = 0\\ \to 4{t^2} – 8t + 3 = 0\end{array}\) Reply
Đáp án:
A
Giải thích các bước giải:
\(\begin{array}{l}
\cos \left( {\dfrac{\pi }{6} – x} \right) = \sin \left( {\dfrac{\pi }{2} – \dfrac{\pi }{6} + x} \right)\\
\to \cos \left( {\dfrac{\pi }{6} – x} \right) = \sin \left( {\dfrac{\pi }{3} + x} \right)\\
Có:\cos 2\left( {x + \dfrac{\pi }{3}} \right) = 1 – 2{\sin ^2}\left( {x + \dfrac{\pi }{3}} \right)\\
Pt \to 1 – 2{\sin ^2}\left( {x + \dfrac{\pi }{3}} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = \dfrac{5}{2}\\
\to 1 – 2.{\cos ^2}\left( {\dfrac{\pi }{6} – x} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) = \dfrac{5}{2}\\
\to – 2.{\cos ^2}\left( {\dfrac{\pi }{6} – x} \right) + 4\cos \left( {\dfrac{\pi }{6} – x} \right) – \dfrac{3}{2} = 0\\
Đặt:\cos \left( {\dfrac{\pi }{6} – x} \right) = t\\
Pt \to – 2{t^2} + 4t – \dfrac{3}{2} = 0\\
\to – 4{t^2} + 8t – 3 = 0\\
\to 4{t^2} – 8t + 3 = 0
\end{array}\)