6$\sqrt[]{2}$sin(x+45)-5sin2x-3=0. Giải nghiệm giúp 19/09/2021 Bởi Gabriella 6$\sqrt[]{2}$sin(x+45)-5sin2x-3=0. Giải nghiệm giúp
Đáp án: \(\left[ \begin{array}{l} x = \alpha – \frac{\pi }{4} + k2\pi \\ x = \frac{{3\pi }}{4} – \alpha + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)\) Giải thích các bước giải: \(\begin{array}{l} 6\sqrt 2 \sin \left( {x + {{45}^0}} \right) – 5\sin 2x – 3 = 0\\ \Leftrightarrow 6\left( {\sin x + \cos x} \right) – 5\sin 2x – 3 = 0\,\,\,\,\left( * \right)\\ Dat\,\,\,\sin x + \cos x = t\,\,\left( { – \sqrt 2 \le t \le \sqrt 2 } \right)\\ \Rightarrow {t^2} = 1 + 2\sin x\cos x = 1 + \sin 2x\\ \Rightarrow \sin 2x = {t^2} – 1\\ \Rightarrow \left( * \right) \Leftrightarrow 6t – 5\left( {{t^2} – 1} \right) – 3 = 0\\ \Leftrightarrow 6t – 5{t^2} + 5 – 3 = 0\\ \Leftrightarrow 5{t^2} – 6t – 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l} t = \frac{{3 + \sqrt {19} }}{5}\,\,\,\left( {ktm} \right)\\ t = \frac{{3 – \sqrt {19} }}{5}\,\,\left( {tm} \right) \end{array} \right.\\ \Rightarrow \sin x + \cos x = \frac{{3 – \sqrt {19} }}{5}\\ \Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{3 – \sqrt {19} }}{5}\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{3 – \sqrt {19} }}{{5\sqrt 2 }} = \sin \alpha \\ \Leftrightarrow \left[ \begin{array}{l} x + \frac{\pi }{4} = \alpha + k2\pi \\ x + \frac{\pi }{4} = \pi – \alpha + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \alpha – \frac{\pi }{4} + k2\pi \\ x = \frac{{3\pi }}{4} – \alpha + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right). \end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
x = \alpha – \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} – \alpha + k2\pi
\end{array} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
6\sqrt 2 \sin \left( {x + {{45}^0}} \right) – 5\sin 2x – 3 = 0\\
\Leftrightarrow 6\left( {\sin x + \cos x} \right) – 5\sin 2x – 3 = 0\,\,\,\,\left( * \right)\\
Dat\,\,\,\sin x + \cos x = t\,\,\left( { – \sqrt 2 \le t \le \sqrt 2 } \right)\\
\Rightarrow {t^2} = 1 + 2\sin x\cos x = 1 + \sin 2x\\
\Rightarrow \sin 2x = {t^2} – 1\\
\Rightarrow \left( * \right) \Leftrightarrow 6t – 5\left( {{t^2} – 1} \right) – 3 = 0\\
\Leftrightarrow 6t – 5{t^2} + 5 – 3 = 0\\
\Leftrightarrow 5{t^2} – 6t – 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = \frac{{3 + \sqrt {19} }}{5}\,\,\,\left( {ktm} \right)\\
t = \frac{{3 – \sqrt {19} }}{5}\,\,\left( {tm} \right)
\end{array} \right.\\
\Rightarrow \sin x + \cos x = \frac{{3 – \sqrt {19} }}{5}\\
\Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{3 – \sqrt {19} }}{5}\\
\Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \frac{{3 – \sqrt {19} }}{{5\sqrt 2 }} = \sin \alpha \\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{4} = \alpha + k2\pi \\
x + \frac{\pi }{4} = \pi – \alpha + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \alpha – \frac{\pi }{4} + k2\pi \\
x = \frac{{3\pi }}{4} – \alpha + k2\pi
\end{array} \right.\,\,\left( {k \in Z} \right).
\end{array}\)