a/

\(\Leftrightarrow1+cos2x+cos3x+cosx=0\)

\(\Leftrightarrow2cos^2x+2cos2x.cosx=0\)

\(\Leftrightarrow2cosx\left(cosx+cos2x\right)=0\)

\(\Leftrightarrow2cosx\left(2cos^2x+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

b/

\(\Leftrightarrow2sin3x.cosx+sin3x=2cos3x.cosx+cos3x\)

\(\Leftrightarrow sin3x\left(2cosx+1\right)-cos3x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left(sin3x-cos3x\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(3x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

c/

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=1-cos4x\)

\(\Leftrightarrow cos6x+cos2x-2cos4x=0\)

\(\Leftrightarrow2cos4x.cos2x-2cos4x=0\)

\(\Leftrightarrow2cos4x\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)

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)

\(2sin^2x+2sin3x.cos2x=0\)

\(\Leftrightarrow sin^2x+\left(3sinx-4sin^3x\right)cos2x=0\)

\(\Leftrightarrow sinx\left[sinx+\left(3-4sin^2x\right)cos2x\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Leftrightarrow...\\sinx+\left(3-4sin^2x\right)cos2x=0\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow sinx+\left(3-4sin^2x\right)\left(1-2sin^2x\right)=0\)

\(\Leftrightarrow8sin^4x-10sin^2x+sinx+3=0\)

\(\Leftrightarrow\left(sinx+1\right)\left(8sin^3x-8sin^2x-2sinx+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\Leftrightarrow...\\8sin^3x-8sin^2x-2sinx+3=0\left(2\right)\end{matrix}\right.\)

Pt (2) có dạng \(8t^3-8t^2-2t+3=0\) là 1 pt bậc 3 ko giải được theo chương trình phổ thông (nghiệm xấu) nên chắc chắn ko thể tìm được nghiệm còn lại của pt

Dù sao giải thì cứ giải chứ 90% là đề bài ko đúng, đề bài là: \(2sin^2x+sinx+sin5x=1\) hợp lý hơn nhiều

Bằng 1 thì ez r cậu :<

a) pt <=> - cos2x. tan²2x + 3.cos2x=0

      <=>  \(\dfrac{sin^22x}{-cos2x}\)+ 3cos2x =0

      <=>  sin²2x - 3cos²2x = 0

     <=> 1 - 4 cos²2x = 0

      <=> 1 - 4.\(\dfrac{1+cos4x}{2}\)= 0

      <=>  cos4x = \(\dfrac{-1}{2}\)