Giải phương trình: 4sin^22x - 3 căn bậc hai 3 sin 4x - 2xos^2 2x = 4
Câu hỏi:
Giải phương trình: 4sin22x – 3\(\sqrt 3 \)sin4x – 2cos22x = 4.
Trả lời:
4sin22x – 3\(\sqrt 3 \)sin4x – 2cos22x = 4
⇔ 2(1 – cos 4x) – 3\(\sqrt 3 \)sin4x – (1 + cos4x) – 4 = 0
⇔ – 3\(\sqrt 3 \)sin4x – 3cos4x = 3
⇔\(\sqrt 3 \)sin4x + cos4x = –1
⇔ \(\frac{{\sqrt 3 }}{2}\)sin4x + \(\frac{1}{2}\)cos4x = \(\frac{{ - 1}}{2}\)
⇔ \(\sin 4x.\cos \left( {\frac{\pi }{6}} \right) + \cos 4x.\sin \left( {\frac{\pi }{6}} \right) = \frac{{ - 1}}{2}\)
⇔ \(\sin \left( {2x + \frac{\pi }{6}} \right) = \frac{{ - 1}}{2}\)
⇔ \(\left[ \begin{array}{l}2x + \frac{\pi }{6} = - \frac{\pi }{6} + k2\pi \\2x + \frac{\pi }{6} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
⇔\(\left[ \begin{array}{l}x = - \frac{\pi }{6} + k\pi \\x = \frac{\pi }{2} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)
