Ta có: sinx/2-cosx/2=1/2

<=> (sinx/2-cosx/2)²=1/4

<=> 1- sinx= 1/4

<=> sinx = 3/4

=> cosx = căn7/4 hoặc cosx= -căn7/4

=> sin2x = 2sinx.cosx

=> sin2x = 3. căn7/8 hoặc sin2x=-3.căn7/8

Đặt \(cosx-sinx=t\Rightarrow-\sqrt{2}\le t\le\sqrt{2}\)

\(t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

Pt trở thành:

\(t\left(1+\dfrac{1-t^2}{2}\right)+1=0\)

\(\Leftrightarrow t^3-3t-2=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=-1\end{matrix}\right.\)

\(\Rightarrow cosx-sinx=-1\)

\(\Leftrightarrow\sqrt[]{2}cos\left(x+\dfrac{\pi}{4}\right)=-1\)

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow...\)

ĐKXĐ:

a. \(cosx\ne0\Leftrightarrow x\ne\frac{\pi}{2}+k\pi\)

b. \(sinx\ne1\Leftrightarrow x\ne\frac{\pi}{2}+k2\pi\)

c. Hàm luôn xác định với mọi x

Đề bài yêu cầu là gì bạn? Bạn không nêu yêu cầu thì làm sao biết để giúp ạ?

Lời giải:
$A=\cos 2x-2\sin 5x\sin x=\cos 2x-2.\frac{-1}{2}[\cos (5x+x)-\cos (5x-x)]$

$=\cos 2x+\cos 6x-\cos 4x$

$=(\cos 2x+\cos 6x)-\cos 4x$

$=2\cos \frac{2x+6x}{2}\cos \frac{6x-2x}{2}-\cos 4x$

$=2\cos 4x\cos 2x-\cos 4x$

$=\cos 4x[2\cos 2x-1]$

Những đáp án A,B,C,D bạn đưa ra không có đáp án nào đúng cả.

những câu hỏi không liên quan đến THCS thì bạn vào h để có thể được giải đáp tốt hơn

"h" nhé mình đánh thiếu 

d/

\(\Leftrightarrow sin2x=sin6x-sin4x\)

\(\Leftrightarrow2sinx.cosx=2cos5x.sinx\)

\(\Leftrightarrow sinx\left(cosx-cos5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos5x=cosx\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\)

a/ Bạn coi lại vế trái đề bài, nhìn không hợp lý

b/ \(\Leftrightarrow\frac{1}{2}sin9x-\frac{1}{2}sinx=\frac{1}{2}sin5x-\frac{1}{2}sinx\)

\(\Leftrightarrow sin9x=sin5x\)

\(\Leftrightarrow\left[{}\begin{matrix}9x=5x+k2\pi\\9x=\pi-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k\pi}{7}\end{matrix}\right.\)

c/ \(\Leftrightarrow sin2x-cos2x=cosx-sinx\)

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

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

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

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

a/

\(\Leftrightarrow2sinx.cosx+2\sqrt{3}cos^2x=\sqrt{3}-2sin5x\)

\(\Leftrightarrow sin2x+\sqrt{3}\left(cos2x+1\right)=\sqrt{3}-2sin5x\)

\(\Leftrightarrow sin2x+\sqrt{3}cos2x=-2sin5x\)

\(\Leftrightarrow\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x=-sin5x\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=sin\left(-5x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{21}+\frac{k2\pi}{7}\\x=-\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)

b/

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

\(\Leftrightarrow sinx-\sqrt{3}cosx=-2sin3x\)

\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=-sin3x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)