Đáp án:
Câu a nhé!
Giải thích các bước giải:
a) \(\dfrac{1}{{\sin \,x}} + \dfrac{1}{{\sin \,\left( {x - \dfrac{{3\pi }}{2}} \right)}} = 4\sin \left( {\dfrac{{7\pi }}{4} - x} \right)\)
\( \Leftrightarrow \dfrac{1}{{\sin \,x}} + \dfrac{1}{{\cos \,x}} = 4\sin \left( {x - \dfrac{{3\pi }}{4}} \right)\) (điều kiện : \(\sin 2x \ne 0\))
\( \Leftrightarrow \dfrac{1}{{\sin \,x}} + \dfrac{1}{{\cos \,x}} = 4\cos \left( {x - \dfrac{\pi }{4}} \right)\)
\( \Leftrightarrow \dfrac{{\sin \,x + \cos \,x}}{{\sin \,x.cos\,x}} = 2\sqrt 2 \left( {\sin \,x + \cos \,x} \right)\)
\( \Leftrightarrow \left[ \begin{array}{l}\sin \,x + \cos x = 0\\2\sqrt 2 \sin \,x.cos\,x = 1\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}\cos \left( {x - \dfrac{\pi }{4}} \right) = 0\\\sin \,2x = \dfrac{1}{{\sqrt 2 }}\end{array} \right.\)
\( \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k2\pi \\x = \dfrac{\pi }{8} + k\pi \\x = \dfrac{{3\pi }}{8} + k\pi \end{array} \right.\) (thỏa mãn)
