(sin²x - 4cos²x)(sin²x - 2sinx.cosx) = 2cos⁴x

⇔ (5sin²x - 4)(sin²x - sin2x) = 2cos⁴x
⇔ \(\left(\dfrac{5-5cos2x}{2}-4\right)\left(\dfrac{1-cos2x}{2}-sin2x\right)\)= 2cos⁴x

⇔ \(\dfrac{5-5cos2x-8}{2}.\dfrac{1-cos2x-2sin2x}{2}\) = 2cos⁴x

⇔ (5cos2x + 3)(cos2x + 2sin2x - 1) = 8cos⁴x

⇔ 5cos²2x + 5cos2x.sin2x + 3cos2x + 6sin2x - 3 = 8cos⁴x

⇔ 5.\(\dfrac{1+cos4x}{2}\) + \(\dfrac{5}{2}sin4x\) + 3cos2x + 6sin2x - 3 = 8cos⁴x

⇔ \(\dfrac{5}{2}cos4x+\dfrac{5}{2}sin4x+3cos2x+6sin2x-\dfrac{1}{2}\) = 8cos⁴x

⇔ 5cos4x + 5sin4x + 6cos2x + 12sin2x - 1 = 16cos⁴x

VP = 16cos⁴x = 16 . \(\dfrac{\left(1+cos2x\right)^2}{4}\) = 4. (1 + cos2x)²

VP = 4 . (1 + 2cos2x + cos²2x)

VP = 4 + 8cos2x + 4 . \(\dfrac{1+cos4x}{2}\)

VP = 6 + 8cos2x+ 2cos4x

Vậy 3cos4x + 5sin4x - 2cos2x + 12sin2x - 7 = 0

@Nguyễn Việt Lâm giúp em với

ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)

\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)

\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)

\(\Leftrightarrow2cos^22x-2cos2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)

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

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

Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow4sin4x.cos4x=\sqrt{2}\)

\(\Leftrightarrow2sin8x=\sqrt{2}\)

\(\Leftrightarrow sin8x=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}8x=\frac{\pi}{4}+k2\pi\\8x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{32}+\frac{k\pi}{4}\\x=\frac{3\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)

a, \(sin4x.cosx-sin3x=0\)

\(\Leftrightarrow\dfrac{1}{2}sin5x+\dfrac{1}{2}sin3x-sin3x=0\)

\(\Leftrightarrow sin5x=sin3x\)

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

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

b, \(sin2x+\sqrt{3}cos2x=\sqrt{2}\)

\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow sinx\left(1-2sin^2x\right)+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)

\(\Leftrightarrow sinx.cos2x+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)

\(\Leftrightarrow sin3x+\sqrt{3}cos3x=2cos4x\)

\(\Leftrightarrow\frac{1}{2}sin3x+\frac{\sqrt{3}}{3}cos3x=cos4x\)

\(\Leftrightarrow sin\left(3x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{2}-4x\right)\)

\(\Leftrightarrow...\)

ĐKXĐ:...

Biến đổi đoạn trong ngoặc trước cho đỡ rối:

\(cos4x+sin2x=cos\left(3x+x\right)+sin\left(3x-x\right)\)

\(=cos3x.cosx-sin3x.sinx+sin3x.cosx-cos3x.sinx\)

\(=cosx\left(cos3x+sin3x\right)-sinx\left(cos3x+sin3x\right)\)

\(=\left(cosx-sinx\right)\left(cos3x+sin3x\right)\)

Thay vào phương trình:

\(\left(cosx-sinx\right)^2=2\left(sinx+cosx\right)+3\)

\(\Leftrightarrow1-2sinx.cosx=2\left(sinx+cosx\right)+3\)

Đặt \(sinx+cosx=a\Rightarrow-2sinx.cosx=1-a^2\)

\(2-a^2=2a+3\Rightarrow a=-1\Rightarrow sinx+cosx=-1\Rightarrow...\)

sin 2   x   -   cos 2   x   =   cos 4 x   ⇔   - cos 2 x   =   cos 4 x   ⇔   2 cos 3 x . cos x   =   0