căn 2 ( cosx+sinx) = 1/2sin2x ai làm ơn giúp mk với 14/09/2021 Bởi Rose căn 2 ( cosx+sinx) = 1/2sin2x ai làm ơn giúp mk với
$\begin{array}{l} Dat\,\sin x + \cos x = t\left( {0 \le t \le \sqrt 2 } \right)\\ \Rightarrow {t^2} = 1 + 2\sin x\cos x = 1 + \sin 2x \Rightarrow \sin 2x = {t^2} – 1\\ \Rightarrow \sqrt 2 t = \frac{1}{2}.\left( {{t^2} – 1} \right) \Leftrightarrow {t^2} – 2\sqrt 2 t – 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = \sqrt 2 + \sqrt 3 \left( {loai} \right)\\ t = \sqrt 2 – \sqrt 3 \left( {TM} \right) \end{array} \right. \Rightarrow \sin x + \cos x = \sqrt 2 – \sqrt 3 \\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \sqrt 2 – \sqrt 3 \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }}\\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} + k2\pi \\ x + \frac{\pi }{4} = \pi – \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} – \frac{\pi }{4} + k2\pi \\ x = \frac{{3\pi }}{4} – \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} + k2\pi \end{array} \right. \end{array}$ Bình luận
$\begin{array}{l}
Dat\,\sin x + \cos x = t\left( {0 \le t \le \sqrt 2 } \right)\\
\Rightarrow {t^2} = 1 + 2\sin x\cos x = 1 + \sin 2x \Rightarrow \sin 2x = {t^2} – 1\\
\Rightarrow \sqrt 2 t = \frac{1}{2}.\left( {{t^2} – 1} \right) \Leftrightarrow {t^2} – 2\sqrt 2 t – 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = \sqrt 2 + \sqrt 3 \left( {loai} \right)\\
t = \sqrt 2 – \sqrt 3 \left( {TM} \right)
\end{array} \right. \Rightarrow \sin x + \cos x = \sqrt 2 – \sqrt 3 \\
\Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \sqrt 2 – \sqrt 3 \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{4} = \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} + k2\pi \\
x + \frac{\pi }{4} = \pi – \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} – \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} – \arcsin \frac{{\sqrt 2 – \sqrt 3 }}{{\sqrt 2 }} + k2\pi
\end{array} \right.
\end{array}$