Question

sin3x + 1=2sin²2x

sin2xcos3x = sin5x

cos5x + cos3x + sin2x =0

sin5x + 1 = 2sin²x

sin3xcosx + 2cos²2x = 1 + cos3xsinx

Answer 1

sin3x + 1=2sin²2x

<=> sin3x + 1 = 2\(\dfrac{1-cos4x}{2}\)

<=> sin3x + 1 = 1 - cos4x

<=> sin3x = -cos4x

<=> sin3x + cos4x = 0

<=> \(\dfrac{\sqrt{2}}{2}\)sin3x + \(\dfrac{\sqrt{2}}{2}\)cos4x = 0 (chia 2 vế cho \(\sqrt{2}\)).

<=> cos\(\dfrac{\pi}{4}\)sin3x + sin\(\dfrac{\pi}{4}\)cos4x = 0

<=> sin (3x+\(\dfrac{\pi}{4}\)) = 0

<=> sin(3x+\(\dfrac{\pi}{4}\)) = sin0

<=> \(\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=0+k2\pi\\3x+\dfrac{\pi}{4}=\pi-0+k2\pi\end{matrix}\right.\)(k\(\in\)Z)

<=>\(\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(k\(\in\)Z)