Giải phương trình: 4(sin^4x + cos^4x) + căn bậc hai 3 sin 4x = 2
Câu hỏi:
Giải phương trình: 4(sin4x + cos4x) + \(\sqrt 3 \sin 4x = 2\).
Trả lời:
4(sin4x + cos4x) + \(\sqrt 3 \sin 4x = 2\)
⇔ 4[(sin2x + cos2x) – 2.sin2x.cos2x + \(\sqrt 3 \sin 4x = 2\)
⇔ \(4\left( {1 - \frac{1}{2}{{\sin }^2}2x} \right) + \sqrt 3 \sin 4x = 2\)
⇔ \(4\left[ {1 - \frac{1}{2}.\left( {\frac{1}{2} - \frac{1}{2}\cos 4x} \right)} \right] + \sqrt 3 \sin 4x = 2\)
⇔ \(4.\frac{3}{4} + 4.\frac{1}{4}.\cos 4x + \sqrt 3 \sin 4x = 2\)
⇔ cos4x + \(\sqrt 3 \sin 4x = - 1\)
⇔ \(\frac{1}{2}\cos 4x + \frac{{\sqrt 3 }}{2}\sin 4x = \frac{{ - 1}}{2}\)
⇔ \(\sin \left( {\frac{\pi }{6}} \right)\cos 4x + \cos \left( {\frac{\pi }{6}} \right)\sin 4x = \frac{{ - 1}}{2}\)
⇔ \(\sin \left( {4x + \frac{\pi }{6}} \right) = \frac{{ - 1}}{2}\)
⇔ \(\left[ \begin{array}{l}4x + \frac{\pi }{6} = \frac{{ - \pi }}{6} + k2\pi \\4x + \frac{\pi }{6} = \pi + \frac{\pi }{6} + k2\pi \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = \frac{{ - \pi }}{{12}} + k\frac{\pi }{2}\\x = \frac{\pi }{4} + k\frac{\pi }{2}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).
