a)\(pt\Leftrightarrow\frac{1-cos8x}{2}+\frac{1-cos6x}{2}=\frac{1-cos4x}{2}+\frac{1-cos2x}{2}\)

\(\Leftrightarrow cos2x+cos4x=cos6x+cos8x\)

\(\Leftrightarrow2cos3x\cdot cosx=2cos7x\cdot cosx\)

\(\Leftrightarrow2cos\left(cos3x-cos7x\right)=0\)

\(\Leftrightarrow2cosx\cdot\left(-2\right)\cdot sin5x\cdot sin\left(-2x\right)=0\)

\(\Leftrightarrow cosx\cdot sin2x\cdot sin5x=0\)

\(\Leftrightarrow sin2x\cdot sin5x=0\)(do sin2x=0 <=>2sinx*cosx=0 gồm th cosx=0 r`)

\(\Leftrightarrow\left[\begin{array}{nghiempt}sin2x=0\\sin5x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{k\pi}{2}\\x=\frac{k\pi}{5}\end{array}\right.\)\(\left(k\in Z\right)\)

b)\(pt\Leftrightarrow1-cos2x+1-cos4x=1+cos6x+1+cos8x\)

\(\Leftrightarrow cos2x+cos8x+cos4x+cos6x=0\)

\(\Leftrightarrow cos10x\cdot cos6x+cos10x\cdot cos2x=0\)

\(\Leftrightarrow cos10x\left(cos6x+cos2x\right)=0\)

\(\Leftrightarrow cos10x\cdot cos8x\cdot cos4x=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}cos10x=0\\cos8x=0\\cos4x=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{16}+\frac{k\pi}{8}\\x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)

\(\left\{{}\begin{matrix}cos^24x+cos^26x\le2\\sin^212x+sin^216x\ge0\end{matrix}\right.\)

\(\Rightarrow VT\le VP\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}cos^24x=1\\cos^26x=1\\sin^212x=0\\sin^216x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin4x=0\\sin6x=0\\sin12x=0\\sin16x=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

a/

\(\Leftrightarrow sinx.cosx\left(sin^2x-cos^2x\right)=\frac{\sqrt{2}}{8}\)

\(\Leftrightarrow2sinx.cosx\left(cos^2x-sin^2x\right)=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow sin2x.cos2x=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow\frac{1}{2}sin4x=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow sin4x=-\frac{\sqrt{2}}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{16}+\frac{k\pi}{2}\\x=\frac{5\pi}{16}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

Câu này đề hơi kì quái, bạn coi lại đề được ko? Biến đổi mấy cách vẫn thấy ko ổn

c/

\(\Leftrightarrow\left(2sinx-cosx+1\right)\left(1+cosx\right)=1-cos^2x\)

\(\Leftrightarrow\left(2sinx-cosx+1\right)\left(1+cosx\right)=\left(1-cosx\right)\left(1+cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}1+cosx=0\left(1\right)\\2sinx-cosx+1=1-cosx\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cosx=-1\Leftrightarrow\pi x=\pi+k2\pi\)

\(\left(2\right)\Leftrightarrow2sinx=0\Rightarrow sinx=0\)

\(\Rightarrow x=k\pi\)

Kết hợp lại ta được \(x=k\pi\)

d/

\(\Leftrightarrow2sin8x.cosx=cos\left(\frac{\pi}{2}-2x\right)+1-1-cos\left(\frac{\pi}{2}+4x\right)\) (hạ bậc vế phải)

\(\Leftrightarrow2sin8x.cosx=sin2x+sin4x\)

\(\Leftrightarrow2sin8x.cosx=2sin3x.cosx\)

\(\Leftrightarrow cosx\left(sin8x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin8x=sin3x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=3x+k2\pi\\8x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=\frac{1}{2}+\frac{1}{2}cos4x+\frac{1}{2}+\frac{1}{2}cos8x\)

\(\Leftrightarrow cos8x+cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2cos5x.cos3x+2cos5x.cosx=0\)

\(\Leftrightarrow cos5x\left(cos3x+cosx\right)=0\)

\(\Leftrightarrow2cos5x.cos2x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos5x=0\\cos2x=0\\cosx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k\pi}{5}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

a: \(\Leftrightarrow sin\left(\dfrac{x}{3}-\dfrac{pi}{4}\right)=sinx\)

=>x/3-pi/4=x+k2pi hoặc x/3-pi/4=pi-x+k2pi

=>2/3x=-pi/4+k2pi hoặc 4/3x=5/4pi+k2pi

=>x=-3/8pi+k3pi hoặc x=15/16pi+k*3/2pi

b: =>(sin3x-sin2x)(sin3x+sin2x)=0

=>sin3x-sin2x=0 hoặc sin 3x+sin 2x=0

=>sin 3x=sin 2x hoặc sin 3x=sin(-2x)

=>3x=2x+k2pi hoặc 3x=pi-2x+k2pi hoặc 3x=-2x+k2pi hoặc 3x=pi+2x+k2pi

=>x=k2pi hoặc x=pi/5+k2pi/5 hoặc x=k2pi/5 hoặc x=pi+k2pi