Giải hộ em vs a nhanh hộ em vs em đang cần gấp lắm ạ . nghiệm của pt (2+căn 3)^cosx + (2-căn3)^cosx=4

$(2 + \sqrt{3})^{cosx} + (2 - \sqrt{3})^{cosx} = 4$

$\Leftrightarrow (2 +\sqrt{3})^{cosx} - (2 + \sqrt{3}) + (2-\sqrt{3})^{cosx} - (2 -\sqrt{3}) = 0$

$\Leftrightarrow (2 + \sqrt{3})\left[(2 + \sqrt{3})^{cosx - 1} - 1\right] = -(2 - \sqrt{3})\left[(2 - \sqrt{3})^{cosx-1} - 1\right]$

$\Leftrightarrow ln\left\{(2 + \sqrt{3})\left[(2 + \sqrt{3})^{cosx - 1} - 1\right]\right\}= ln\left\{-(2 - \sqrt{3})\left[(2 - \sqrt{3})^{cosx-1} - 1\right]\right\}$

$\Leftrightarrow ln(2 + \sqrt{3}) + ln\left[(2 + \sqrt{3})^{cosx - 1} - 1\right] = -ln(2 - \sqrt{3}) + ln\left[(2 - \sqrt{3})^{cosx - 1} - 1\right]$

$\Leftrightarrow \left[ln(2 + \sqrt{3}) + ln(2 - \sqrt{3})\right] + ln\left[(2 + \sqrt{3})^{cosx - 1} - 1\right] = ln\left[(2 - \sqrt{3})^{cosx - 1} - 1\right]$

$\Leftrightarrow ln\left[(2 + \sqrt{3})^{cosx - 1} - 1\right] = ln\left[(2 - \sqrt{3})^{cosx - 1} - 1\right]$

$\Leftrightarrow (2 + \sqrt{3})^{cosx-1} = (2 - \sqrt{3})^{cosx - 1}$

$\Leftrightarrow (cosx-1)ln(2 + \sqrt{3}) = (cosx - 1)ln(2 -\sqrt{3}) = 0$

$\Leftrightarrow (cosx - 1)ln\left(\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}\right) = 0$

$\Leftrightarrow cosx = 1$

$\Leftrightarrow x = k2\pi \,\, (k \in \Bbb Z)$