a) 2cos2x - 3cosx + 1 = 0 (1)
Đặt cosx = t vớit ∈ [-1 ; 1]
(1) <=> 2t2 - 3t + 1 = 0 => t = 1 hoặc t = 1/2
Với t = 1 => cosx = 1 => $x = k2\pi , k \epsilon Z$
Với t = 1/2 => $cosx = \frac{1}{2} \Leftrightarrow x = \pm \frac{\pi }{3} + k2\pi , k \epsilon Z$
b) 2sin2x + $\sqrt{2}$sin4x = 0
$\Leftrightarrow 2sin2x(1 + \sqrt{2}cos2x)=0$
$\Leftrightarrow\left[ \begin{array}{ll} sin2x = 0 \\ 1 + \sqrt{2}cos2x = 0\end{array} \right.$
$\Leftrightarrow\left[ \begin{array}{ll} sin2x = 0 \\ cos2x = - \frac{1}{\sqrt{2}}\end{array} \right.$
$\Leftrightarrow\left[ \begin{array}{ll} 2x = k\pi \\ 2x = \pm \frac{3\pi}{4} + k2\pi\end{array} \right.$
$\Leftrightarrow\left[ \begin{array}{ll} x = k\frac{\pi}{2} , k \epsilon Z \\ x = \pm \frac{3\pi}{8} + k\pi, k \epsilon Z \end{array} \right.$
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