a) √2 cos(x - π/4)

= √2.(cosx.cos π/4 + sinx.sin π/4)

= √2.(√2/2.cosx + √2/2.sinx)

= √2.√2/2.cosx + √2.√2/2.sinx

= cosx + sinx (đpcm)

b) √2.sin(x - π/4)

= √2.(sinx.cos π/4 - sin π/4.cosx )

= √2.(√2/2.sinx - √2/2.cosx )

= √2.√2/2.sinx - √2.√2/2.cosx

= sinx – cosx (đpcm).

pt <=> 1+cos2x + cos3x + cosx = 0

<=> 2cos²x + 2cos2x.cosx = 0 

<=> 2cosx.(cos2x + cosx) = 0 
<=> 4cosx.cos(3x/2).cos(x/2) = 0 <=> 
[cosx = 0 
[cos(3x/2) = 0 (tập nghiệm cos3x/2 = 0 chứa tập nghiệm cosx/2 = 0) 
<=> 
[x = pi/2 + kpi 
[3x/2 = pi/2 + kpi 
<=> 
[x = pi/2 + kpi 
[x = pi/3 + 2kpi/3 (k thuộc Z) 

sin^2 x + sin^2 2x + sin^2 3x + sin^2 4x = 
[1-cos(2x)]/2+ [1-cos(4x)]/2+[1-cos(6x)]/2+[1-cos(8x)]/... = 
2- [ cos(2x)+cos(4x)+cos(6x)+cos(8x)]/2 = 
2- 1/2· [ cos(2x)+cos(8x)]+cos(4x)+cos(6x)]= 
2- 1/2· [ 2·cos(-3x)·cos(5x) + 2· cos(-x)·cos(5x)]= 
2- cos(5x)· [cos(3x)+cosx] = 
2- cos(5x)· 2·cos(2x)·cosx = 
2- 2·cosx·cos(2x)·cos(5x)= 2 <--> 

*cosx=0 --> x= pi/2+ k·pi with k thuộc Z or 
*cos(2x)=0 --> x= pi/4 + k·pi/2 with k thuộc Z or 
* cos(5x)=0 --> x= pi/10+ k·pi/5 with k thuộc Z 

2 bài kia chịu  cùy :))

1.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos4x=\dfrac{1}{2}+\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow-cos4x=cos\left(2x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow cos\left(4x-\pi\right)=cos\left(2x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\pi=2x-\dfrac{\pi}{2}+k2\pi\\4x-\pi=\dfrac{\pi}{2}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{3}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{3}\)

2.

\(\Leftrightarrow1-cos^2x+1-sin^24x=2\)

\(\Leftrightarrow cos^2x+sin^24x=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx=0\\sin4x=0\end{matrix}\right.\)

\(\Leftrightarrow cosx=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)