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

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

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 $x=\dfrac{-\pi}{6}+k2\pi$ ($k\in\mathbb Z$).