Câu hỏi:

Ta có phương trình: $2\cos \left( {x + \frac{\pi }{4}} \right) - \sqrt 2 = 0$

$2\cos \left( {x + \frac{\pi }{4}} \right) = \sqrt 2 $

$\cos \left( {x + \frac{\pi }{4}} \right) = \frac{\sqrt 2 }{2}$

$\cos \left( {x + \frac{\pi }{4}} \right) = \cos \frac{\pi }{4}$

$x + \frac{\pi }{4} = \pm \frac{\pi }{4} + k2\pi ,k \in \mathbb{Z}$

$\left[ \begin{array}{l}x = - \frac{\pi }{4} + \frac{\pi }{4} + k2\pi \\x = - \frac{\pi }{4} - \frac{\pi }{4} + k2\pi \end{array} \right.,k \in \mathbb{Z}$

$\left[ \begin{array}{l}x = k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.,k \in \mathbb{Z}$

Ta có: $\sin(x + \frac{\pi}{3}) + \sin(2x) = 0$
$\Leftrightarrow 2\sin(\frac{3x + \frac{\pi}{3}}{2})\cos(\frac{-x + \frac{\pi}{3}}{2}) = 0$
$\Leftrightarrow \left[ \begin{array}{l}\sin(\frac{3x}{2} + \frac{\pi}{6}) = 0 \\ \cos(\frac{-x}{2} + \frac{\pi}{6}) = 0 \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}\frac{3x}{2} + \frac{\pi}{6} = k\pi \\ \frac{-x}{2} + \frac{\pi}{6} = \frac{\pi}{2} + k\pi \end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x = -\frac{\pi}{9} + k\frac{2\pi}{3} \\ x = -\frac{2\pi}{3} - 2k\pi \end{array} \right. (k \in \mathbb{Z})$

Ta cần giải phương trình $v(t) = 3$, tức là:
$3\sin \left( {2t - \frac{\pi }{6}} \right) = 3$
$\Leftrightarrow \sin \left( {2t - \frac{\pi }{6}} \right) = 1$
$\Leftrightarrow 2t - \frac{\pi }{6} = \frac{\pi }{2} + k2\pi, k \in \mathbb{Z}$
$\Leftrightarrow 2t = \frac{\pi }{2} + \frac{\pi }{6} + k2\pi$
$\Leftrightarrow 2t = \frac{{4\pi }}{6} + k2\pi$
$\Leftrightarrow t = \frac{{2\pi }}{6} + k\pi = \frac{\pi }{3} + k\pi, k \in \mathbb{Z}$
Với $k = 1$, ta có $t = \frac{\pi }{3} + \pi = \frac{{4\pi }}{3}$.

Ta có phương trình ${2^x} = 5$.
Áp dụng định nghĩa logarit, ta có $x = \log_2 5$.