Verified answer

[tex]\dfrac{\sin3x}{\sin x} + \dfrac{\cos3x}{\cos x}=\dfrac{\sin3x\cos x+\cos3x\sin x}{\sin x\cos x}=\dfrac{\sin(3x+x)}{\sin x\cos x}=\dfrac{\sin4x}{\sin x\cos x}=[/tex]

[tex]=\dfrac{2\sin4x}{2\sin x\cos x}=\dfrac{2\sin4x}{\sin2x}=\dfrac{2\cdot2\sin2x\cos2x}{\sin2x}=\dfrac{4\sin2x\cos2x}{\sin2x}=4\cos2x[/tex]

При [tex]x=\dfrac{\pi }{6}[/tex]:

[tex]4\cos\left(2\cdot\dfrac{\pi }{6}\right)=4\cos\dfrac{\pi }{3}=4\cdot\dfrac{1}{2}=2[/tex]

Ответ: 2