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

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

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

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

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

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.

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.

Chọn A

y = cos6 x+ sin2xcos2x(sin2x + cos2x) + sin4x - sin2x

= cos6x + sin2x(1 - sin2x) + sin4x - sin2x = cos6x

Do đó : y' = -6cos5xsinx.

\(D=\frac{1+sin2x+cos2x}{1+sin2x-cos2x}=\frac{1+2sinxcosx+2cos^2x-1}{1+2sinxcosx-1+2sin^2x}\)

\(D=\frac{cosx\left(sinx+cosx\right)}{sinx\left(sinx+cosx\right)}=cotx\)

\(F=\frac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)

\(F=\frac{2sin4xcos3x+sin4x}{2cos4xcos3x+cos4x}\)

\(F=\frac{2sin4x\left(cos3x+1\right)}{2cos4x\left(cos3x+1\right)}=tan4x\)

\(\Leftrightarrow sin4x\left(sin5x+sin3x\right)-sin2x.sinx=0\)

\(\Leftrightarrow2sin^24x.cosx-2sin^2x.cosx=0\)

\(\Leftrightarrow cosx\left(2sin^24x-2sin^2x\right)=0\)

\(\Leftrightarrow cosx\left(1-cos8x-1+cos2x\right)=0\)

\(\Leftrightarrow cosx\left(cos2x-cos8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=2x+k2\pi\\8x=-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{k\pi}{3}\\x=\frac{k\pi}{5}\end{matrix}\right.\)

\(sin ⁡ 5 x - sin ⁡ 4 x + sin ⁡ 3 x \left(\right. 1 \left.\right)\)

\(\Leftrightarrow \left(\right. sin ⁡ 3 x + sin ⁡ 5 x \left.\right) - sin ⁡ 4 x = 0\)

\(\Leftrightarrow 2 sin ⁡ 4 x . cos ⁡ x - sin ⁡ 4 x = 0\)

\(\Leftrightarrow sin ⁡ 4 x \left(\right. 2 cos ⁡ x - 1 \left.\right) = 0\)

\(\Leftrightarrow \left[\right. sin ⁡ 4 x = 0 \\ 2 cos ⁡ x - 1 = 0 \Leftrightarrow \left[\right. x = \frac{k \pi}{4} \\ x = \frac{\pi}{3} + k 2 \pi \\ x = - \frac{\pi}{3} + k 2 \pi\)\(\left(\right. k \in \mathbb{Z} \left.\right)\)

Vậy các nghiệm của phương trình là \(x = \frac{k \pi}{4}\), \(x = \frac{\pi}{3} + k 2 \pi\) và  \(x = - \frac{\pi}{3} + k 2 \pi\) \(\left(\right. k \in \mathbb{Z} \left.\right)\).

\(\cos^4x-\sin^4x=\cos^4x-\left(sin^2x.sin^2x\right)=\cos^4x-\left(1-cos^2x\right)\left(1-cos^2x\right)\)

=\(2cos^2x-1=2cos^2x-sin^2x-cos^2x=cos^2x-sin^2x\)

bn làm vế trái hay phải vậy