Đáp án:
$\left[ \begin{array}{l}x=\dfrac{\pi}{12}-\dfrac{∝-k2\pi}{3}\\x=\dfrac{\pi}{4}-\dfrac{∝-k2\pi}{3}\end{array} \right.$ ($k∈\mathbb{Z}$)
Với $\left\{ \begin{array}{l}sin∝=\dfrac{5}{\sqrt[]{26}}\\cos∝=\dfrac{1}{\sqrt[]{26}}\end{array} \right.$
Giải thích các bước giải:
$5cos3x+sin3x=\sqrt[]{13}$
$↔ \dfrac{5}{\sqrt[]{26}}cos3x+\dfrac{1}{\sqrt[]{26}}sin3x=\dfrac{1}{\sqrt[]{2}}$
Đặt $\left\{ \begin{array}{l}sin∝=\dfrac{5}{\sqrt[]{26}}\\cos∝=\dfrac{1}{\sqrt[]{26}}\end{array} \right.$, ta có:
$sin∝.cos3x+cos∝.sin3x=\dfrac{1}{\sqrt[]{2}}$
$↔ sin(∝+3x)=sin\Bigg(\dfrac{\pi}{4}\Bigg)$
$↔ \left[ \begin{array}{l}∝+3x=\dfrac{\pi}{4}+k2\pi\\∝+3x=\dfrac{3\pi}{4}+k2\pi\end{array} \right.$
$↔ \left[ \begin{array}{l}x=\dfrac{\pi}{12}-\dfrac{∝-k2\pi}{3}\\x=\dfrac{\pi}{4}-\dfrac{∝-k2\pi}{3}\end{array} \right.$ ($k∈\mathbb{Z}$)
