\(sin^2x+cosx.cos3x+sin2x.cos2x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x+\dfrac{1}{2}sin4x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}-sin^2x+\dfrac{1}{2}sin4x+\dfrac{1}{2}cos4x=0\)

\(\Leftrightarrow sin4x+cos4x=-1\)

\(\Leftrightarrow\sqrt{2}sin\left(4x+\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow sin\left(4x+\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\4x+\dfrac{\pi}{4}=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Đặt \(t=\sin2x+cos2x\)

=>\(t^2=\left(\sin2x+cos2x\right)^2=1+2\cdot\sin2x\cdot cos2x\)

=>\(2\cdot\sin2x\cdot cos2x=t^2-1\)

=>\(\sin2x\cdot cos2x=\frac{t^2-1}{2}\)

Ta có: \(t=\sin2x+cos2x\)

=>\(t=\sqrt2\cdot\sin\left(2x+\frac{\pi}{4}\right)\)

Ta có: \(-1<=\sin\left(2x+\frac{\pi}{4}\right)\le1\)

=>\(-\sqrt2\le\sqrt2\cdot\sin\left(2x+\frac{\pi}{4}\right)\le\sqrt2\)

=>\(-\sqrt2\le t\le\sqrt2\)

Ta có: \(\left(\sin2x+cos2x\right)\left(1-\sin2x\cdot cos2x\right)+\sin2x\cdot cos2x=1\)

=>\(t\cdot\left(1-\frac{t^2-1}{2}\right)+\frac{t^2-1}{2}=1\)

=>\(t\cdot\frac{2-t^2+1}{2}+\frac{t^2-1}{2}=1\)

=>\(\frac{-t^3+3t+t^2-1}{2}=1\)

=>\(-t^3+t^2+3t-1=2\)

=>\(t^3-t^2-3t+1=-2\)

=>\(t^3-t^2-3t+3=0\)

=>\(\left(t-1\right)\left(t^2-3\right)=0\)

=>\(\left[\begin{array}{l}t-1=0\\ t^2-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}t=1\\ t^2=3\end{array}\right.\Rightarrow\left[\begin{array}{l}t=1\left(nhận\right)\\ t=\sqrt3\left(loại\right)\\ t=-\sqrt3\left(loại\right)\end{array}\right.\)

=>t=1

=>\(\sqrt2\cdot\sin\left(2x+\frac{\pi}{4}\right)=1\)

=>\(\sin\left(2x+\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\ 2x+\frac{\pi}{4}=\pi-\frac{\pi}{4}+k2\pi=\frac34\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=k2\pi\\ 2x=\frac12\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=k\pi\\ x=\frac{\pi}{4}+k\pi\end{array}\right.\)

\(VT=sin^4x\cdot\dfrac{cos^2x}{sin^2x}+cos^4x\cdot\dfrac{sin^2x}{cos^2x}+sin^4x-sin^2x\cdot cos^2x\)

\(=sin^2x\cdot cos^2x+cos^2x\cdot sin^2x+sin^4x-sin^2x\cdot cos^2x\)

\(=sin^2x\left(sin^2x+cos^2x\right)=sin^2x=VP\)

\(\sin4x=2\sin2x.\cos2x\)

\(\Rightarrow\sin2x.\cos2x=\frac{1}{2}\sin4x\)

\(-1\le\sin4x\le1\)

\(\Rightarrow\frac{-1}{2}\le\frac{1}{2}\sin4x\le\frac{1}{2}\Rightarrow\left\{{}\begin{matrix}y_{max}=\frac{1}{2};"="\Leftrightarrow x=\frac{\pi}{2}+k2\pi\\y_{min}=-\frac{1}{2};"="\Leftrightarrow x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

Chọn C.

Ta có

C = [ ( sin²x + cos²x) – sin²cos²x]² - [ ( sin⁴x + cos⁴x) ² - 2sin⁴x.cos⁴x]

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

= 2[ 1-sin²x.cos²x]² - [1-sin²x.cos²x]² + 2sin⁴x.cos⁴x

= 2( 1 - 2sin²x.cos²x + sin⁴x.cos⁴x)- ( 1 - 4sin²xcos²x + 4sin⁴x.cos⁴x) + 2sin⁴x.cos⁴x

= 1.

a/ ĐKXĐ: \(cosx\ne-\frac{1}{2}\)

\(\Leftrightarrow2cosx-1=6cosx+3\)

\(\Leftrightarrow4cosx=-4\Rightarrow cosx=-1\)

\(\Rightarrow x=\pi+k2\pi\)

b/

\(\Leftrightarrow cosx\left(2cos2x-1\right)-3cosx=0\)

\(\Leftrightarrow cosx\left(2cos2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=2\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\frac{\pi}{2}+k\pi\)

c/

\(\Leftrightarrow2sin2x.cos2x=0\)

\(\Leftrightarrow sin4x=0\)

\(\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}\)

Chọn C.

Ta có: C = 2( sin⁴x + cos⁴x + sin²x.cos²x) ² - ( sin⁸x + cos⁸x)

= 2 [ (sin²x + cos²x) ² - sin²x.cos²x]² - [ (sin⁴x + cos⁴x)² - 2sin⁴x.cos⁴x]

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

= 2[ 1- sin²x.cos²x]² - [ 1 - 2sin²x.cos²x]²  + 2sin⁴x.cos⁴x

= 2( 1 - 2sin²xcos²x+ sin⁴x.cos⁴x) –( 1- 4sin²xcos²x+ 4sin⁴xcos⁴x) + 2sin⁴x.cos⁴x

=  1.