Chọn A.

Ta có:

+ sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x.cos²x = 1 - 2sin²x.cos²x.

+ sin⁴x + cos⁴x = 1 - 3sin²x.cos²x.

Do đó

A = 3(1 - 2sin²x^.cos²x) - 2(1 - 3sin²x.cos²x) = 1.

\(\Leftrightarrow\sin^2x-\sin^22x+\sin^23x-\sin^24x=0\)

\(\Leftrightarrow\left(\sin x+\sin2x\right)\left(\sin2x-\sin x\right)+\left(\sin3x+\sin4x\right)\left(\sin4x-\sin3x\right)=0\)

\(\Leftrightarrow2\sin\dfrac{3}{2}x.\cos\dfrac{x}{2}.2\cos\dfrac{3}{2}x.\sin\dfrac{x}{2}+2\sin\dfrac{7}{2}x.\cos\dfrac{x}{2}.2\sin\dfrac{x}{2}\cos\dfrac{7}{2}x=0\)

\(\Leftrightarrow\sin3x.\sin x+\sin7x.\sin x=0\)

\(\Leftrightarrow\sin x\left(\sin3x+\sin7x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin x=0\\\sin3x=\sin\left(-7x\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\3x=-7x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)

Chứng minh các biểu thức đã cho không phụ thuộc vào x.

f(x) = 1 ⇒ f′(x) = 0

\(A=3\left[\left(sin^2x+cos^2x\right)^2-2\cdot sin^2x\cdot cos^2x\right]-2\left[\left(sin^2x+cos^2x\right)^3-3\cdot sin^2x\cdot cos^2x\left(sin^2x+cos^2x\right)\right]\)
 

\(=3\left[1-2\cdot sin^2x\cdot cos^2x\right]-2\left[1-3\cdot sin^2x\cdot cos^2x\right]\)

\(=3-6\cdot sin^2x\cdot cos^2x-2+6\cdot sin^2x\cdot cos^2x\)

=1

A =  cos 6 x  + 3 sin 2 x . cos 2 x  + 2 sin 4 α .  cos 2 x  +  sin 4 α

=  cos 6 x  + 3.(1 -  cos 2 x ) cos 4 x  + 2 sin 4 α . cos 2 x  +  sin 4 α

=  cos 6 x  + 3 cos 4 x  - 3 cos 6 x  + 2 sin 4 α  - 2 sin 6 x  +  sin 4 α

= -2.( cos 6 x  +  sin 6 x ) + 3  cos 4 x  + 3 sin 4 α

= -2.( cos 6 x  +  sin 6 x ) + 3.( cos 4 x  +  sin 4 α ) = 1

Vậy biểu thức A không phụ thuộc vào x.

\(pt\Leftrightarrow cos6x+3cos2x-4\left(2cos^2x-1\right)=0\)

\(\Leftrightarrow cos6x+3cos2x-4cos2x=0\)

\(\Leftrightarrow cos6x-cos2x=0\)

\(\Leftrightarrow-2sin4x.sin2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin2x=0\end{matrix}\right.\)

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

Đáp án B