Bài 1:

ĐK : sinx cosx > 0

Khi đó phương trình trở thành

sinx+cosx=\(2\sqrt{\sin x\cos x}\)

ĐK sinx + cosx >0 → sinx>0 ; cosx>0

Khi đó \(2\sqrt{\sin x\cos x}\Leftrightarrow2\sin x=1\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

Vậy ...

Bài 2:

ĐK : \(\sin\left(3x+\frac{\pi}{4}\right)\ge0\)

Khi đó phương trình đã cho tương đương với phương trình \(\sin2x=\frac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

Trong khoảng từ \(\left(-\pi,\pi\right)\) ta nhận được các giá trị :

\(x=\frac{\pi}{12}\) (TMĐK)

\(x=-\frac{11\pi}{12}\) (KTMĐK)

\(x=\frac{5\pi}{12}\) (KTMĐK)

\(x=-\frac{7\pi}{12}\) (TMĐK)

Vậy ta có 2 nghiệm thõa mãn \(x=\frac{\pi}{12}\) và \(x=-\frac{7\pi}{12}\)

\(\Leftrightarrow\cos4x+\cos2x+\sqrt{3}\left(1+\sin2x\right)=\sqrt{3}\left(1+\cos\left(4x+\frac{\pi}{2}\right)\right)\)

\(\Leftrightarrow\cos4x+\sqrt{3}\sin4x+\sqrt{3}\sin2x=0\)

\(\Leftrightarrow\sin\left(4x+\frac{\pi}{6}\right)+\sin\left(2x+\frac{\pi}{6}\right)=0\)

\(\Leftrightarrow2\sin\left(3x+\frac{\pi}{6}\right)\cos x=0\)

\(\Leftrightarrow\begin{cases}x=-\frac{\pi}{18}+k\frac{\pi}{3}\\x=\frac{\pi}{2}+k\pi\end{cases}\)

Vậy phương trình có 2 nghiệm \(x=-\frac{\pi}{18}+k\frac{\pi}{3}\) và \(x=\frac{\pi}{2}+k\pi\)

a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp

b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)

\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)

\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)

\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)

c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:

\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)

Đặt \(\sqrt{tanx+1}=t\ge0\)

\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)

\(\Leftrightarrow3t^3-5t^2+3t-10=0\)

\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)

d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

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

Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)

\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)

\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)

a) \(\cos \left( {3x - \frac{\pi }{4}} \right) =  - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} =  - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi  + k2\pi }\\{3x =  - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)

\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x =  - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} =  - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} =  - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x =  - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi  + k2\pi }\\{x =  - \pi  + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)

c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x =  - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)

3.

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

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

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

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

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm

a)      

\(\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) =  - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi  + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

b)     \(\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

c)       

\(\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x =  - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x =  - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x =  - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}\)

d)      

\(\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x =  - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x =  - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}\)

e)      

\(\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}\)

f)       

\(\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x =  - \frac{\pi }{4} + k\pi ;k \in Z\end{array}\)

\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt[]{3}sin2x-cos2x\)

\(\Leftrightarrow4.\left(-\dfrac{1}{2}\right)\left[sin\left(x+\dfrac{\pi}{3}+x-\dfrac{\pi}{6}\right)+sin\left(x+\dfrac{\pi}{3}-x+\dfrac{\pi}{6}\right)\right]=m^2+2.\left[\dfrac{\sqrt[]{3}}{2}.sin2x-\dfrac{1}{2}.cos2x\right]\)

\(\Leftrightarrow2\left[sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(2x-\dfrac{\pi}{6}\right)\right]=m^2+2\)

\(\Leftrightarrow2.2sin2x.cos\dfrac{\pi}{6}=m^2+2\)

\(\Leftrightarrow2.2sin2x.\dfrac{\sqrt[]{3}}{2}=m^2+2\)

\(\Leftrightarrow2\sqrt[]{3}sin2x.=m^2+2\)

\(\Leftrightarrow sin2x.=\dfrac{m^2+2}{2\sqrt[]{3}}\)

Phương trình có nghiệm khi và chỉ khi

\(\left|\dfrac{m^2+2}{2\sqrt[]{3}}\right|\le1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m^2+2}{2\sqrt[]{3}}\ge-1\\\dfrac{m^2+2}{2\sqrt[]{3}}\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2\ge-2\left(1+\sqrt[]{3}\right)\left(luôn.đúng\right)\\m^2\le2\left(1-\sqrt[]{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow-\sqrt[]{2\left(1-\sqrt[]{3}\right)}\le m\le\sqrt[]{2\left(1-\sqrt[]{3}\right)}\)

a/ \(cosx-cos2x+sin2x-sinx=3-4cosx\)

\(\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0\)

\(\Leftrightarrow sinx\left(2cosx-1\right)-\left(2cosx-1\right)\left(cosx-2\right)=0\)

\(\Leftrightarrow\left(2cosx-1\right)\left(sinx-cosx+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)

b/ ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.\) \(\Rightarrow...\)

\(\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}\)

\(\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0\)

\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx-\sqrt{3}\right)+\sqrt{3}sinx\left(2cosx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx+\sqrt{3}sinx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)

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)