2.1

a.

\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)

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

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

b.

\(cosx-\sqrt{3}sinx=1\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)

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

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

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

\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)

\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Chọn D

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k2\pi\\x\ne-\dfrac{\pi}{6}+k2\pi\\x\ne\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\dfrac{cosx-2sinx.cosx}{1-2sin^2x+sinx}=\sqrt{3}\)

\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)

\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

ĐKXĐ : \(sinx\ne1;-\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+2k\pi\\x\ne\dfrac{-\pi}{6}+2k\pi;\dfrac{7\pi}{6}+2k\pi\end{matrix}\right.\)   

\(\Leftrightarrow x\ne\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\)( k thuộc Z ) 

P/t đã cho \(\Leftrightarrow\dfrac{cosx-sin2x}{1-2sin^2x+sinx}=\sqrt{3}\) 

\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(cos2x+sinx\right)\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

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

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

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

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

Vậy ...

√3tan⁡x + 1 = 0 ⇔ tan⁡x = (-√3)/3 ⇔ x = (-π)/6 + kπ, k ∈ Z)

Lời giải:
PT $\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=2\sin x\cos x-\sin x$

$\Leftrightarrow (2\cos x-1)(2\sin x+\cos x)=\sin x(2\cos x-1)$

$\Leftrightarrow (2\cos x-1)(\sin x+\cos x)=0$

$\Rightarrow 2\cos x=1$ hoặc $\sin x=-\cos x=\cos (\pi -x)=\sin (x-\frac{\pi}{2})$

Đến đây thì đơn giản rồi.

Nkjuiopmli Sv5: Bạn chuyển vế sin x(2cos x-1) sang vế trái thì vế phải còn 0 đó.