Sinx+cosx.sin2x+(căn3)cos3x=2(cos4x+sin^3x)

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1. Đáp án: $x=\dfrac{\pi}{42}+k\dfrac{2\pi}{7}$ và $x=\dfrac{-\pi}{6}+k2\pi$ ($k\in\mathbb Z$)

    

   Giải thích các bước giải:

   $\Rightarrow \sin x+\cos x\sin 2x+\sqrt3\cos 3x=2(\cos4x+\sin 3x)$

   $\Rightarrow \sin x+2{\cos}^2x\sin x-2{\sin}^3x+\sqrt3\cos 3x=2\cos 4x$

   $\Rightarrow \sin x+2\sin x({\cos}^2x-{\sin}^2x)+\sqrt3\cos 3x=2\cos 4x$

   $\Rightarrow \sin x+2\sin x\cos 2x+\sqrt3\cos 3x=2\cos 4x$

   $\Rightarrow \sin x+2.\dfrac{1}{2}[\sin 3x+\sin(-x)]+\sqrt3\cos 3x=2\cos 4x$

   $\Rightarrow \sin 3x+\sqrt3\cos 3x=2\cos 4x$

   $\Rightarrow \dfrac{1}{2}\sin 3x+\dfrac{\sqrt3}{2}\cos 3x=\cos 4x$

   $\Rightarrow \sin(3x+\dfrac{\pi}{3})=\sin(\dfrac{\pi}{2}-4x)$

   $\Rightarrow 3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}-4x+k2\pi$ hoặc $3x+\dfrac{\pi}{3}=\pi-(\dfrac{\pi}{2}-4x)+k2\pi$

   Vậy $x=\dfrac{\pi}{42}+k\dfrac{2\pi}{7}$ và $x=\dfrac{-\pi}{6}+k2\pi$ ($k\in\mathbb Z$).

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2. sinx + cosx.sin2x + √3.cos3x = 2(cos4x + sin³x)

   <=> sinx + 2cos²x.sinx – 2sin³x + √3.cos3x = 2cos4x

   <=> sinx + 2sinx.(cos²x – sin²x) + √3.cos3x = 2cos4x

   <=> sinx + 2sinx.cos2x + √3.cos3x = 2cos4x

   <=> sinx + sin3x + sin(-x) + √3.cos3x = 2cos4x

   <=> (1/2).sin3x + (√3/2).cos3x = cos4x

   <=> sin(pi/6).sin3x + cos(pi/6).cos3x = cos4x

   <=> cos(3x-pi/6) = cos4x

   <=> [ 4x = 3x-pi/6 + 2kpi

   —— [ 4x = -3x+pi/6 + 2kpi

   <=> [ x = -pi/6 + 2kpi

   —— [ x = pi/42 + 2kpi/7

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