Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} \\
= \sqrt {{{\sin }^4}x + 4.\left( {1 - {{\sin }^2}x} \right)} + \sqrt {{{\cos }^4}x + 4.\left( {1 - {{\cos }^2}x} \right)} \\
= \sqrt {{{\sin }^4}x - 4{{\sin }^2}x + 4} + \sqrt {{{\cos }^4}x - 4{{\cos }^2}x + 4} \\
= \sqrt {{{\left( {{{\sin }^2}x - 2} \right)}^2}} + \sqrt {{{\left( {{{\cos }^2}x - 2} \right)}^2}} \\
= \left( {2 - {{\sin }^2}x} \right) + \left( {2 - {{\cos }^2}x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {0 \le {{\sin }^2}x,{{\cos }^2}x \le 1} \right)\\
= 4 - \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
= 4 - 1 = 3\\
3\tan \left( {x + \dfrac{\pi }{3}} \right).\tan \left( {\dfrac{\pi }{6} - x} \right)\\
= 3.\dfrac{{\sin \left( {x + \dfrac{\pi }{3}} \right)}}{{\cos \left( {x + \dfrac{\pi }{3}} \right)}}.\dfrac{{\sin \left( {\dfrac{\pi }{6} - x} \right)}}{{\cos \left( {\dfrac{\pi }{6} - x} \right)}}\\
= 3.\dfrac{{\sin \left( {x + \dfrac{\pi }{3}} \right)}}{{\cos \left( {x + \dfrac{\pi }{3}} \right)}}.\dfrac{{\cos \left( {\dfrac{\pi }{2} - \dfrac{\pi }{6} + x} \right)}}{{\sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{6} + x} \right)}}\\
= 3.\dfrac{{\sin \left( {x + \dfrac{\pi }{3}} \right)}}{{\cos \left( {x + \dfrac{\pi }{3}} \right)}}.\dfrac{{\cos \left( {x + \dfrac{\pi }{3}} \right)}}{{\sin \left( {x + \dfrac{\pi }{3}} \right)}}\\
= 3\\
\Rightarrow \sqrt {{{\sin }^4}x + 4{{\cos }^2}x} + \sqrt {{{\cos }^4}x + 4{{\sin }^2}x} = 3\tan \left( {x + \dfrac{\pi }{3}} \right).\tan \left( {\dfrac{\pi }{6} - x} \right)
\end{array}\)
