2sinx(1+cos2x)+sin2x=1+2cosx

Đáp án:

$\left[\begin{array}{l}x = \pm \dfrac{2\pi}{3} + k2\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right.\quad (k \in \Bbb Z)$

Giải thích các bước giải:

$\begin{array}{l}2\sin x(1 + \cos2x) + \sin2x = 1 + 2\cos x\\ \Leftrightarrow 2\sin x(1 + 2\cos^2x - 1) + \sin2x = 1 + 2\cos x\\ \Leftrightarrow 2\sin x.2\cos^2x + \sin2x = 1 + 2\cos x\\ \Leftrightarrow 2\sin2x.\cos x+ \sin2x = 1 + 2\cos x\\ \Leftrightarrow \sin2x(2\cos x +1) =1+2\cos x\\ \Leftrightarrow (2\cos x+1)(\sin2x - 1) =0\\ \Leftrightarrow \left[\begin{array}{l}2\cos x + 1 = 0\\\sin2x - 1 =0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\cos x = -\dfrac12\\\sin2x = 1\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \pm \dfrac{2\pi}{3} + k2\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$

`2sin x(1 + cos 2x) + sin 2x = 1 + 2cos x`

`-> 2sin x.(1 + 2cos^2 x - 1) + sin 2x - (1 + 2cos x) = 0`

`-> 2sin x.cos x.2cos x + sin 2x - (1 + 2cos x) = 0`

`-> 2.sin 2x.cos x + sin 2x - (1 + 2cos x) = 0`

`-> sin 2x(2cos x + 1) - (1 + 2cos x) = 0`

`-> (2cos x + 1)(sin 2x - 1) = 0`

`->` \(\left[ \begin{array}{l}cos x = -\dfrac{1}{2}\\sin 2x = 1\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x = ±\dfrac{2\pi}{3} + k2\pi\\x = \dfrac{\pi}{4} + kπ\end{array} \right.\) `(k in ZZ)`