1)pt\(\Leftrightarrow sin^8x\left(1-2sin^2x\right)=cos^8x\left(2cos^2x-1\right)+\frac{5}{4}cos2x\)

\(\Leftrightarrow sin^8x.cos2x=cos^8x.cos2x+\frac{5}{4}cos2x\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}cos2x=0\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin^8x-cos^8x=\frac{5}{4}\left(\cdot\right)\end{array}\right.\)

Xét (*):VT(*)\(\le sin^8x\le1\)\(\Rightarrow\)pt(*) vô ngiệm

Vậy pt có 1 họ nghiệm là \(x=\frac{\pi}{4}+\frac{k\pi}{2},k\in Z\)

2)+)sinx=0 không là nghiệm của pt

+)sinx\(\ne0\):

pt\(\Leftrightarrow16sinx.cosx.cos2x.cos4x.cos8x=1\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=1\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=1\)\(\Leftrightarrow2sin8x.cos8x=1\Leftrightarrow sin16x=1\Leftrightarrow x=\frac{\pi}{32}+\frac{k\pi}{8},k\in Z\)

KL:...

\(\Leftrightarrow\left(sin^4x+cos^4x\right)^2-2sin^4x.cos^4x=\frac{17}{32}\)

\(\Leftrightarrow\left[1-2sin^2x.cos^2x\right]^2-2sin^4x.cos^4x=\frac{17}{32}\)

Đặt \(sin^2x.cos^2x=\frac{1}{4}sin^22x=t\Rightarrow0\le t\le\frac{1}{4}\)

\(\Rightarrow\left(1-2t\right)^2-2t^2=\frac{17}{32}\)

\(\Leftrightarrow2t^2-4t+\frac{15}{32}=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{15}{8}\left(l\right)\\t=\frac{1}{8}\end{matrix}\right.\) \(\Rightarrow\frac{1}{4}sin^22x=\frac{1}{8}\Leftrightarrow2sin^22x=1\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow sin8x-\sqrt{2}cos8x=cos6x-\sqrt{2}sin6x\)

\(\Leftrightarrow\dfrac{1}{\sqrt{3}}sin8x-\dfrac{\sqrt{2}}{\sqrt{3}}cos8x=\dfrac{1}{\sqrt{3}}cos6x-\dfrac{\sqrt{2}}{\sqrt{3}}sin6x\)

Đặt \(\dfrac{1}{\sqrt{3}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{\sqrt{2}}{\sqrt{3}}=sina\)

\(\Rightarrow sin8x.cosa-cos8x.sina=cos6x.cosa-sin6x.sina\)

\(\Leftrightarrow sin\left(8x-a\right)=cos\left(6x+a\right)\)

\(\Leftrightarrow sin\left(8x-a\right)=sin\left(\dfrac{\pi}{2}-6x-a\right)\)

\(\Leftrightarrow...\)

a.

\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{1}{2}\)

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

\(\Leftrightarrow cos^2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

c.

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

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow3-3sin^22x=4cos^22x\)

\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)

\(\Leftrightarrow3=3+cos^22x\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)

2.\(sin^22x+cos^23x=1\)

\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)

\(\Leftrightarrow cos6x=cos4x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)

Vậy...

3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)

\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)

\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))

Vậy...

4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)

\(\Leftrightarrow cos2x+cos4x=1+cos6x\)

\(\Leftrightarrow2cos3x.cosx=2cos^23x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...