Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + {\cos ^2}x = 1 \Rightarrow \sin x = \pm \sqrt {1 - {{\cos }^2}x} = \pm \dfrac{{\sqrt {15} }}{4}\\
\sin 2x = 2\sin x.\cos x = \pm \dfrac{{\sqrt {15} }}{8}\\
\cos 2x = 2{\cos ^2}x - 1 = - \dfrac{7}{8}\\
\tan \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{{\sin \left( {2x + \dfrac{\pi }{4}} \right)}}{{\cos \left( {2x + \dfrac{\pi }{4}} \right)}}\\
= \dfrac{{\sin 2x.\cos \dfrac{\pi }{4} + \cos 2x.\sin \dfrac{\pi }{4}}}{{\cos 2x.\cos \dfrac{\pi }{4} - \sin 2x.\sin \dfrac{\pi }{4}}}\\
= \dfrac{{\sin 2x + \cos 2x}}{{\cos 2x - \sin 2x}}\,\,\,\,\,\,\,\,\,\left( {\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{{\sqrt 2 }}{2}} \right)\\
*)\\
\sin 2x = \dfrac{{\sqrt {15} }}{8} \Rightarrow \tan \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{{32 - 7\sqrt {15} }}{{17}}\\
*)\\
\sin 2x = - \dfrac{{\sqrt {15} }}{8} \Rightarrow \tan \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{{32 + 7\sqrt {15} }}{{17}}
\end{array}\)
