Giải pt:
$\sin (2x - \dfrac{\pi }{4}) = \sin (x - \dfrac{\pi }{4}) + \dfrac{{\sqrt 2 }}{2}$
Mọi người làm cho vui!^^
Started By Hưng 11a1 ĐTĐ, 15-09-2009 - 22:29
#1
Posted 15-09-2009 - 22:29
#2
Posted 16-09-2009 - 07:45
$\sin (2x - \dfrac{\pi }{4}) = \sin (x - \dfrac{\pi }{4}) + \dfrac{{\sqrt 2 }}{2}$Giải pt:
$\sin (2x - \dfrac{\pi }{4}) = \sin (x - \dfrac{\pi }{4}) + \dfrac{{\sqrt 2 }}{2}$
$\Leftrightarrow \sin (2x - \dfrac{\pi }{4})-sin\dfrac{\pi}{4}=\sin (x - \dfrac{\pi }{4})$
$\Leftrightarrow 2cosx.sin(x-\dfrac{\pi}{4})-sin(x-\dfrac{\pi}{4})=0$
$ \Leftrightarrow \left[ \begin{matrix} sin(x - \dfrac{\pi }{4}) = 0 \\ \cos x = \dfrac{1}{2} = \cos \dfrac{\pi }{3} \\ \end{matrix} \right.$
$ \Leftrightarrow \left[ \begin{matrix} x = \dfrac{\pi }{4} + k\pi \\ x = \pm \dfrac{\pi }{3} + k2\pi \\ \end{matrix} \right.(k \in Z)$
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