Đáp án:
\[A = \tan 2x\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{{\sin \left( {2019\pi - x} \right) + \cos \left( {\dfrac{\pi }{2} - 2x} \right) + \sin 3x}}{{\cos \left( {2020\pi - x} \right) - \cos \left( {\pi + 2x} \right) + \cos 3x}}\\
= \dfrac{{\sin \left( {2.1009\pi + \left( {\pi - x} \right)} \right) + \sin 2x + \sin 3x}}{{\cos \left( { - x} \right) + \cos \left( {\pi - \left( {\pi + 2x} \right)} \right) + \cos 3x}}\\
= \dfrac{{\sin \left( {\pi - x} \right) + \sin 2x + \sin 3x}}{{\cos x + \cos \left( { - 2x} \right) + \cos 3x}}\\
= \dfrac{{\sin x + \sin 2x + \sin 3x}}{{\cos x + \cos 2x + \cos 3x}}\\
= \dfrac{{\left( {\sin x + \sin 3x} \right) + \sin 2x}}{{\left( {\cos x + \cos 3x} \right) + \cos 2x}}\\
= \dfrac{{2.\sin \dfrac{{x + 3x}}{2}.\cos \dfrac{{3x - x}}{2} + \sin 2x}}{{2.cos\dfrac{{x + 3x}}{2}.cos\dfrac{{3x - x}}{2} + \cos 2x}}\\
= \dfrac{{2\sin 2x.\cos x + \sin 2x}}{{2\cos 2x.\cos x + \cos 2x}}\\
= \dfrac{{\sin 2x\left( {2\cos x + 1} \right)}}{{\cos 2x\left( {2\cos x + 1} \right)}}\\
= \dfrac{{\sin 2x}}{{\cos 2x}} = \tan 2x
\end{array}\)
