giải pt 19/ $\frac{(1+sinx+cos2x)sin(x+\frac{ π}{4})}{1+tanx}=\frac{1}{ √2}cosx$

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1. Đáp án:

   $\left[\begin{array}{l} x = \dfrac{\pi}{2} + k2\pi\\x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$

   Giải thích các bước giải:

   $\dfrac{(1 + \sin x + \cos2x)\sin\left(x + \dfrac{\pi}{4}\right)}{1 + \tan x} = \dfrac{1}{\sqrt2}\cos x\qquad (*)$

   $ĐKXĐ: \, \begin{cases}\cos x \ne 0\\\tan x \ne – 1 \end{cases}\Leftrightarrow \begin{cases}x \ne \dfrac{\pi}{2} + n\pi\\x \ne -\dfrac{\pi}{4} + n\pi\end{cases} \quad (n\in \Bbb Z)$

   $(*)\Leftrightarrow \dfrac{(1 + \sin x + \cos2x)\sqrt2\sin\left(x + \dfrac{\pi}{4}\right)}{\dfrac{\sin x + \cos x}{\cos x}} = \cos x$

   $\Leftrightarrow \dfrac{\cos x(1 + \sin x + \cos2x)(\sin x + \cos x)}{\sin x + \cos x} = \cos x$

   $\Leftrightarrow 1 + \sin x + \cos2x = 1 \qquad (\cos x \ne 0)$

   $\Leftrightarrow \sin x + \cos2x = 0$

   $\Leftrightarrow \sin x + 1 – 2\sin^2x = 0$

   $\Leftrightarrow \left[\begin{array}{l}\sin x = 1\\\sin x = -\dfrac{1}{2}\end{array}\right.$

   $\Leftrightarrow \left[\begin{array}{l} x = \dfrac{\pi}{2} + k2\pi\\x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$

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