\(2\sqrt 2 {\rm{ sin }}\left( {x + \frac{\pi }{4}} \right) =
\frac{1}{{{\mathop{\rm s}\nolimits} {\rm{inx}}}} + \frac{1}{{\cos x}} \)\(\Leftrightarrow
2\sqrt 2 {\rm{ sin }}\left( {x + \frac{\pi }{4}} \right) =
\frac{{{\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x}}{{\sin
{\rm{x}}\cos x}}\)\( \Leftrightarrow 2\sqrt 2 {\rm{ sin }}\left( {x +
\frac{\pi }{4}} \right) = \frac{{\sqrt 2 \sin \left( {x + \frac{\pi
}{4}} \right)}}{{\sin {\rm{x}}\cos x}}\) $\Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right)\left[ {2 - \frac{1}{{\sin {\rm{x}}\cos x}}} \right] = 0$$\sin \left( {x + \frac{\pi }{4}} \right)=0\Leftrightarrow x = \pm \frac{\pi }{4} + k\pi {\rm{ }}\left( {k \in Z} \right)$$2
- \frac{1}{{\sin {\rm{x}}\cos x}}=0\Leftrightarrow \sin 2x
=1\Leftrightarrow x = \pm \frac{\pi }{4} + k\pi {\rm{ }}\left( {k \in
Z} \right)$Đáp số: $x = \pm \frac{\pi }{4} + k\pi {\rm{ }}\left( {k \in Z} \right)$
Điều kiện: $\sin x\ne0,\cos x\ne0$ \(2\sqrt 2 {\rm{ sin }}\left( {x + \frac{\pi }{4}} \right) =
\frac{1}{{{\mathop{\rm s}\nolimits} {\rm{inx}}}} + \frac{1}{{\cos x}} \)\(\Leftrightarrow
2\sqrt 2 {\rm{ sin }}\left( {x + \frac{\pi }{4}} \right) =
\frac{{{\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x}}{{\sin
{\rm{x}}\cos x}}\)\( \Leftrightarrow 2\sqrt 2 {\rm{ sin }}\left( {x +
\frac{\pi }{4}} \right) = \frac{{\sqrt 2 \sin \left( {x + \frac{\pi
}{4}} \right)}}{{\sin {\rm{x}}\cos x}}\) $\Leftrightarrow \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right)\left[ {2 - \frac{1}{{\sin {\rm{x}}\cos x}}} \right] = 0$$\sin \left( {x + \frac{\pi }{4}} \right)=0\Leftrightarrow x =
- \frac{\pi }{4} + k\pi {\rm{ }}\left( {k \in
\mathbb{Z
} } \right)$$2
- \frac{1}{{\sin {\rm{x}}\cos x}}=0\Leftrightarrow \sin 2x
=1\Leftrightarrow x = \frac{\pi }{4} + k\pi {\rm{ }}\left( {k \in
\mathbb{Z
} } \right)$Đáp số: $x = \pm \frac{\pi }{4} + k\pi {\rm{ }}\left( {k \in
\mathbb{Z
} } \right)$