Căn3 cos(x – pi/2) + sin(x – pi/2) = 2sin2x 01/10/2021 Bởi Savannah Căn3 cos(x – pi/2) + sin(x – pi/2) = 2sin2x
Đáp án: \(\left\{ \begin{array}{l} x = – \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{7\pi }}{{18}} + k2\pi \end{array} \right.(k \in \mathbb Z)\) Lời giải: \(\begin{array}{l} \sqrt 3 \cos \left( {x – \dfrac{\pi }{2}} \right) + \sin \left( {x – \dfrac{\pi }{2}} \right) = 2\sin 2x\\ \Rightarrow \sqrt 3 \left( {\cos x\cos \dfrac{\pi }{2} + \sin x\sin \dfrac{\pi }{2}} \right) + \left( {\sin x\cos \dfrac{\pi }{2} – \cos x\sin \frac{\pi }{2}} \right) \\= 2\sin 2x\\ \Rightarrow \sqrt 3 \sin x – \cos x = 2\sin 2x\\ \Rightarrow \dfrac{{\sqrt 3 }}{2}\sin x – \dfrac{1}{2}\cos x = \sin 2x\\ \Rightarrow \sin \left( {x – \dfrac{\pi }{6}} \right) = \sin 2x\\ \Rightarrow \left[ \begin{array}{l} x – \dfrac{\pi }{6} = 2x + k2\pi \\ x – \dfrac{\pi }{6} = \pi – 2x + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = – \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{7\pi }}{{18}} + k2\pi \end{array} \right.(k \in \mathbb Z)\\ \end{array}\) Bình luận
Đáp án:
\(\left\{ \begin{array}{l} x = – \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{7\pi }}{{18}} + k2\pi \end{array} \right.(k \in \mathbb Z)\)
Lời giải:
\(\begin{array}{l} \sqrt 3 \cos \left( {x – \dfrac{\pi }{2}} \right) + \sin \left( {x – \dfrac{\pi }{2}} \right) = 2\sin 2x\\ \Rightarrow \sqrt 3 \left( {\cos x\cos \dfrac{\pi }{2} + \sin x\sin \dfrac{\pi }{2}} \right) + \left( {\sin x\cos \dfrac{\pi }{2} – \cos x\sin \frac{\pi }{2}} \right) \\= 2\sin 2x\\ \Rightarrow \sqrt 3 \sin x – \cos x = 2\sin 2x\\ \Rightarrow \dfrac{{\sqrt 3 }}{2}\sin x – \dfrac{1}{2}\cos x = \sin 2x\\ \Rightarrow \sin \left( {x – \dfrac{\pi }{6}} \right) = \sin 2x\\ \Rightarrow \left[ \begin{array}{l} x – \dfrac{\pi }{6} = 2x + k2\pi \\ x – \dfrac{\pi }{6} = \pi – 2x + k2\pi \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = – \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{7\pi }}{{18}} + k2\pi \end{array} \right.(k \in \mathbb Z)\\ \end{array}\)
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