1.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos4x=\dfrac{1}{2}+\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow-cos4x=cos\left(2x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow cos\left(4x-\pi\right)=cos\left(2x-\dfrac{\pi}{2}\right)\)

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

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

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{3}\)

2.

\(\Leftrightarrow1-cos^2x+1-sin^24x=2\)

\(\Leftrightarrow cos^2x+sin^24x=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx=0\\sin4x=0\end{matrix}\right.\)

\(\Leftrightarrow cosx=0\)

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

1.

\(sin\left(4x-10^0\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(4x-10^0\right)=sin45^0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10^0=45^0+k360^0\\4x-10^0=135^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=55^0+k360^0\\4x=145^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=13,75^0+k90^0\\x=36,25^0+k90^0\end{matrix}\right.\) (\(k\in Z\))

2.

Đề không đúng

3.

ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(tan2x=tanx\)

\(\Rightarrow2x=x+k\pi\)

\(\Rightarrow x=k\pi\)

4.

\(cot\left(x+\dfrac{\pi}{5}\right)=-1\)

\(\Leftrightarrow x+\dfrac{\pi}{5}=-\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=-\dfrac{9\pi}{20}+k\pi\) (\(k\in Z\))

5.

\(cos3x=sin5x\)

\(\Leftrightarrow sin5x=sin\left(\dfrac{\pi}{2}-3x\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) (\(k\in Z\))

\(\Leftrightarrow\dfrac{3x}{4}-\dfrac{\pi}{6}=k\pi\)

\(\Leftrightarrow\dfrac{3x}{4}=\dfrac{\pi}{6}+k\pi\)

\(\Leftrightarrow3x=\dfrac{2\pi}{3}+k4\pi\)

\(\Leftrightarrow x=\dfrac{2\pi}{9}+\dfrac{k4\pi}{3}\) (\(k\in Z\))

a) √2 cos(x - π/4)

= √2.(cosx.cos π/4 + sinx.sin π/4)

= √2.(√2/2.cosx + √2/2.sinx)

= √2.√2/2.cosx + √2.√2/2.sinx

= cosx + sinx (đpcm)

b) √2.sin(x - π/4)

= √2.(sinx.cos π/4 - sin π/4.cosx )

= √2.(√2/2.sinx - √2/2.cosx )

= √2.√2/2.sinx - √2.√2/2.cosx

= sinx – cosx (đpcm).

\(sinx+cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)

\(=\sqrt{2}cos\left(\frac{\pi}{2}-\left(x+\frac{\pi}{4}\right)\right)=\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)

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

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

\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x\)

\(=sin^2x-cos^2x+sin2x=sin2x-cos2x\)

\(=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)

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