Chủ đề này có 1 trả lời

[iamshant]

$2cos3x.cosx+\sqrt{3}(1+sin2x)=2\sqrt{3}.cos^{2}(2x+\frac{\pi }{4})$

 

[sieumau88]

$2cos3x.cosx+\sqrt{3}(1+sin2x)=2\sqrt{3}.cos^{2}(2x+\frac{\pi }{4})$

 
$2cos3x.cosx + \sqrt{3} . (1+sin2x) = \sqrt{3} . \left [1+cos\left (4x+\dfrac{\pi }{2}\right )\right ]$
 
$\Leftrightarrow 2cos3x.cosx + \sqrt{3} + \sqrt{3} . sin2x = \sqrt{3} - \sqrt{3} . sin4x$
 
$\Leftrightarrow 2cos3x.cosx + \sqrt{3} . (sin2x + sin4x) = 0$
 
$\Leftrightarrow 2cos3x.cosx + 2\sqrt{3} . sin3x . cosx = 0$
 
$\Leftrightarrow 4cosx . \left (\dfrac{1}{2}cos3x + \dfrac{\sqrt{3}}{2}sin3x \right ) = 0$
 
$\Leftrightarrow 4cosx . \left (sin\dfrac{\pi}{6}cos3x + cos\dfrac{\pi}{6}sin3x \right ) = 0$
 
$\Leftrightarrow 4cosx . sin \left (\dfrac{\pi}{6} + 3x \right ) = 0$
 
$\Leftrightarrow \left[ \begin{array}{l} cosx = 0 \\ sin\left ( \dfrac{\pi}{6} + 3x \right ) = 0 \\ \end{array} \right.$
 
$\rightarrow$ ........v......v........