Giải phương trình (1 + căn bậc hai 3) sin x + (1 - căn bậc hai 3) cos x = 2

\(\left( {1 + \sqrt 3 } \right)\sin x + \left( {1 - \sqrt 3 } \right)\cos x = 2\)

\( \Leftrightarrow \frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}\sin x + \frac{{1 - \sqrt 3 }}{{2\sqrt 2 }}\cos x = \frac{1}{{\sqrt 2 }}\)

Đặt \(\frac{{1 + \sqrt 3 }}{{2\sqrt 2 }} = \cos \alpha \) thì \(\frac{{1 - \sqrt 3 }}{{2\sqrt 2 }} = \sin \alpha \), khi đó PT \( \Leftrightarrow \sin \left( {x + \alpha } \right) = \sin \frac{\pi }{4}\).

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x + \alpha = \frac{\pi }{4} + k2\pi }\\{x + \alpha = \frac{{3\pi }}{4} + k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \alpha + \frac{\pi }{4} + k2\pi }\\{x = - \alpha + \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\) .