Ta có

\(\begin{array}{l}\sin \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{4}{\rm{ }} = {\rm{ }}\frac{\pi }{4} + k2\pi ;k \in Z\\x + \frac{\pi }{4}{\rm{ }} = {\rm{ }}\pi {\rm{ - }}\frac{\pi }{4} + k2\pi ;k \in Z\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = {\rm{ }}k2\pi ;k \in Z\\x{\rm{ }} = {\rm{ }}\frac{\pi }{2} + k2\pi ;k \in Z\end{array} \right.\end{array}\)

Mà \(x \in \left[ {0;\pi } \right]\) nên \(x \in \left\{ {0;\frac{\pi }{2}} \right\}\)

Vậy phương trình đã cho có số nghiệm là 2.

Chọn C

\(\Leftrightarrow2\left(cos^2x-sin^2x\right)+sinx.cosx\left(sinx+cosx\right)=m\left(sinx+cosx\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(\text{vô nghiệm trên đoạn xét}\right)\\2cosx-2sinx+sinx.cosx=m\left(1\right)\end{matrix}\right.\) 

Xét (1), đặt \(t=cosx-sinx=\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}t\in\left[-1;1\right]\\sinx.cosx=\dfrac{1-t^2}{2}\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2t+\dfrac{1-t^2}{2}=m\)

Xét hàm \(f\left(t\right)=-\dfrac{1}{2}t^2+2t+\dfrac{1}{2}\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=2\notin\left[-1;1\right]\) ; \(f\left(-1\right)=-2\) ; \(f\left(1\right)=2\)

\(\Rightarrow-2\le f\left(t\right)\le2\Rightarrow-2\le m\le2\)

Câu 3 sai đề hả bạn

1.

\(\Leftrightarrow2x-\frac{\pi}{4}=x+\frac{\pi}{3}+k\pi\)

\(\Rightarrow x=\frac{7\pi}{12}+k\pi\)

\(-\pi< \frac{7\pi}{12}+k\pi< \pi\Rightarrow-\frac{19}{12}< k< \frac{5}{12}\Rightarrow k=\left\{-1;0\right\}\) có 2 nghiệm

\(x=\left\{-\frac{5\pi}{12};\frac{7\pi}{12}\right\}\)

2.

\(\Leftrightarrow3x-\frac{\pi}{3}=\frac{\pi}{2}+k\pi\)

\(\Rightarrow x=\frac{5\pi}{18}+\frac{k\pi}{3}\)

Nghiệm âm lớn nhất là \(x=-\frac{\pi}{18}\) khi \(k=-1\)

3.

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13\pi}{12}+k2\pi\\x=\frac{17\pi}{12}+k2\pi\end{matrix}\right.\)

Nghiệm âm lớn nhất \(x=-\frac{7\pi}{12}\) ; nghiệm dương nhỏ nhất \(x=\frac{13\pi}{12}\)

Tổng nghiệm: \(\frac{\pi}{2}\)

\(\cos5x=-\sin4x\)

<=> \(\cos5x=\cos\left(4x+\frac{\pi}{2}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}5x=4x+\frac{\pi}{2}+k2\pi\\5x=-4x-\frac{\pi}{2}+k2\pi\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{\pi}{2}+k2\pi\\x=-\frac{\pi}{18}+\frac{k2\pi}{9}\end{cases}}\)

Nghiệm âm lớn nhất: \(-\frac{\pi}{18}\)

Nghiệm dương  nhỏ nhất: \(\frac{\pi}{2}\)

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

<=> \(\sin\left(5x+\frac{\pi}{3}\right)=\sin\left(2x+\frac{\pi}{6}\right)\)

<=> \(\orbr{\begin{cases}5x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\5x+\frac{\pi}{3}=\pi-2x-\frac{\pi}{6}+k2\pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{\pi}{14}+\frac{k2\pi}{7}\end{cases}}\)

Trên \(\left[0,\pi\right]\)có các nghiệm:

\(\frac{11\pi}{18},\frac{\pi}{14},\frac{5\pi}{14},\frac{9\pi}{14},\frac{13\pi}{14}\)

tính tổng:...

Câu 1 với câu 2 sai đề, sin và cos nằm trong [-1;1], mà căn 2 với căn 3 lớn hơn 1 rồi

3/ \(\sin x=\cos2x=\sin\left(\frac{\pi}{2}-2x\right)\)

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

4/ \(\Leftrightarrow\cos^2x-2\sin x\cos x=0\)

Xét \(\cos x=0\) là nghiệm của pt \(\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(\cos x\ne0\Rightarrow1-2\tan x=0\Leftrightarrow\tan x=\frac{1}{2}\Rightarrow x=...\)

5/ \(\Leftrightarrow\sin\left(2x+1\right)=-\cos\left(3x-1\right)=\cos\left(\pi-3x+1\right)=\sin\left(\frac{\pi}{2}-\pi+3x-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\frac{\pi}{2}-\pi+3x-1\\2x+1=\pi-\frac{\pi}{2}+\pi-3x+1\end{matrix}\right.\Leftrightarrow....\)

6/ \(\Leftrightarrow\cos\left(\pi\left(x-\frac{1}{3}\right)\right)=\frac{1}{2}\Leftrightarrow\pi\left(x-\frac{1}{3}\right)=\pm\frac{\pi}{3}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{1}{3}+2k\Rightarrow x=\frac{2}{3}+2k\left(1\right)\\x-\frac{1}{3}=-\frac{1}{3}+2k\Rightarrow x=2k\left(2\right)\end{matrix}\right.\)

\(\left(1\right):-\pi< x< \pi\Rightarrow-\pi< \frac{2}{3}+2k< \pi\) (Ủa đề bài sai hay sao ý nhỉ?)

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

Thui, để đây bao giờ...hết lười thì làm tiếp :(

7)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{42}+k\frac{2\pi}{7}\\x=\frac{\pi}{6}+k\frac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

Do:\(0< x< \pi\)

\(Với:x=\frac{-\pi}{42}+k\frac{2\pi}{7}\left(k\in Z\right)\Rightarrow khôngtìmđượck\)

\(Với:x=\frac{\pi}{6}+k\frac{2\pi}{3}\left(k\in Z\right)\Leftrightarrow\frac{1}{4}< k< \frac{5}{4}\Rightarrow k=\left\{0;1\right\}\Rightarrow\left[{}\begin{matrix}k=0\Rightarrow x=\frac{\pi}{6}\\k=1\Rightarrow x=\frac{5\pi}{6}\end{matrix}\right.\)

Vậy nghiệm của pt là: \(x=\frac{\pi}{6};x=\frac{5\pi}{6}\)

8)