\(P=\dfrac{1+2sin3xcos3x-\left(1-2sin^23x\right)}{1+2sin3xcos3x+2cos^2x-1}=\dfrac{2sin3xcos3x+2sin^23x}{2sin3xcos3x+2cos^23x}=\dfrac{sin3x}{cos3x}=tan3x\)

\(x=\dfrac{7\pi}{4}\Rightarrow P=tan\dfrac{21\pi}{4}=tan\dfrac{\pi}{4}=1\)

Tham khảo (có cả Min lẫn Max):

\(y=\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\)

\(\sin^6x+\cos^6x=\left(\sin^2x+\cos^2x\right)\left(\sin^4x-\sin^2x\cdot\cos^2x+\cos^4x\right)\\ =\left(\sin^2x+\cos^2x\right)^2-3\sin^2x\cdot\cos^2x=1-\dfrac{3}{4}\sin^22x\)

Do \(0\le\sin^22x\le1\Leftrightarrow\dfrac{3}{4}\cdot0\ge-\dfrac{3}{4}\sin^22x\ge-\dfrac{3}{4}\)

\(\Leftrightarrow1\ge1-\dfrac{3}{4}\sin^22x\ge1-\dfrac{3}{4}=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{4}{3}\ge\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)\ge\dfrac{1}{4}\cdot\dfrac{4}{3}=\dfrac{1}{3}\)

Ta có \(-1\le\cos4x\le1\)

\(\Leftrightarrow\dfrac{1}{3}-1-1\le\dfrac{4}{3}\left(\sin^6x+\cos^6x\right)+\cos4x-1\le\dfrac{4}{3}+1-1\\ \Leftrightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)

Vậy \(y_{min}=-\dfrac{5}{3};y_{max}=\dfrac{4}{3}\)

\(y=\dfrac{4}{3}\left(sin^6x+cos^6x\right)+cos4x-1\)

\(y=\dfrac{4}{3}\left(\dfrac{5}{8}+\dfrac{3}{8}cos4x\right)+cos4x-1\)

\(y=\dfrac{3}{2}cos4x-\dfrac{1}{6}\)

\(-1\le cos4x\le1\Rightarrow-\dfrac{5}{3}\le y\le\dfrac{4}{3}\)

\(y_{min}=-\dfrac{5}{3}\) khi \(cos4x=-1\)

\(y_{max}=\dfrac{4}{3}\) khi \(cos4x=1\)

Ủa đề yêu cầu làm gì bạn? Đây ko phải là phương trình

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

Đáp án: C

Ta có:

sin 6 x  +  c o s 6 x  = ( sin 2 x ) 3  + ( cos 2 x ) 3

= ( sin 2 x  +  c o s 2 x )( sin 4 x  -  sin 2 x cos 2 x  +  c o s 4 x )

=  sin 4 x  -  sin 2 x cos 2 x  +  c o s 4 x

= ( sin 2 x  +  cos 2 x ) 2  - 3  sin 2 x cos 2 x

= 1 - 3 sin 2 x cos 2 x

= 1 - (3/4)  sin 2 2 x

Vì

Vậy giá trị nhỏ nhất của  sin 6 x  +  c o s 6 x  là 1/4

Dấu “=” xảy ra ⇔  sin 2 2 x  = 1 ⇔ sin⁡2x = 1 hoặc sin⁡2x = -1