$\lim_{n \to \infty}\dfrac{|a_{n+1}|}{|a_{n}|} = \lim_{n \to \infty}\left [ \dfrac{3\left ( n+1 \right )+1}{4\left ( n+1 \right )+2} \cdot \dfrac{3n+1}{4n+2} \right ] = ...... =  \dfrac{9}{16}$
 

Vậy khoảng hội tụ là_ $\left(\dfrac{-16}{9} ; \dfrac{16}{9}\right)$___ $\Leftrightarrow \dfrac{-16}{9} < x < \dfrac{16}{9}$
 
Khi_ $x=\pm \dfrac{16}{9}$ _, chuỗi đã cho có dạng_ $\sum_{n=1}^{+\infty}(-1)^n \cdot \left(\dfrac{3n+1}{4n+2}\right) \cdot \left(\dfrac{16}{9}\right)^n$

 

Ta có__ $\dfrac{|u_{n+1}|}{|u_{n}|} = \left[\dfrac{3(n+1)+1}{4(n+1)+2} \cdot \left(\dfrac{16}{9}\right)^{n+1}\right] : \left[\dfrac{3n+1}{4n+2} \cdot \left(\dfrac{16}{9}\right)^n\right] = ............ = \dfrac{16}{9} > 1$
 

Vậy__ $\sum_{n=1}^{+\infty}(-1)^n \cdot \left(\dfrac{3n+1}{4n+2}\right) \cdot \left(\dfrac{16}{9}\right)^n$__phân kỳ .
 
Kết luận , miền hội tụ của chuỗi đã cho là_ $\left(\dfrac{-16}{9} ; \dfrac{16}{9}\right)$

Câu 3 :

 

$3 .cos 4x - 2 .cos^{2} 3x = 1$

$\Leftrightarrow 3. cos 4x - \left ( 1 + cos 6x \right ) = 1$

$\Leftrightarrow 3. cos 4x - cos 6x - 2 = 0$

 

Đặt__ $t = cos 2x$

 

pt $\Leftrightarrow 3  \left ( 2t^2 - 1 \right ) - \left ( 4t^3 - 3t \right ) - 2 =0$

 

$\rightarrow$ pt bậc 3 theo t $\rightarrow$ giải tìm t  .......

đề bài là gì vậy bạn ?

Câu 1:      $M=x.(y+z)^{2}+y.(z+x)^{2}+z.(x+y)^{2}-4xyz$
$M=[x.(y+z)^{2}-2xyz]+[y.(z+x)^{2}-2xyz]+z.(x+y)^{2}$
$M=x.(y^2+z^2)+y.(x^2+z^2)+z.(x+y)^{2}$
$M=xy.(x+y)+z^2.(x+y)+z.(x+y)^{2}$
$M=(x+y)(xy+z^2+zx+zy)$
$M=(x+y)(x+y)(y+z)$

$2.sinx.\left ( 1+tanx.tan\dfrac{x}{2} \right )=sin2x+1$
ĐK:    $\left\{\begin{matrix}
cos \ x \neq 0 \\
cos \ \dfrac{x}{2} \neq 0
\end{matrix}\right.$
pt $\Leftrightarrow 2.sinx.\left ( 1+\dfrac{sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}} \right )=sin2x+1$
$\Leftrightarrow 2.sinx.\left ( \dfrac{cosx.cos\frac{x}{2}+sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}} \right )=sin2x+1$
$\Leftrightarrow 2.sinx.\left [ \dfrac{cos\left (x-\frac{x}{2} \right )}{cosx.cos\frac{x}{2}} \right ]=sin2x+1$
$\Leftrightarrow 2.tanx=sin2x+1$

$\Leftrightarrow 2.t = \dfrac{2.t}{t^2+1}+1$        ( Đặt    $t=tanx$)

 

$\Leftrightarrow 2.t+2.t^3 =2.t+t^2+1$

 

$\Leftrightarrow 2.t^3 -t^2-1=0$         (pt này có 1 nghiệm là t=1)

...........v...........v.............

$A=sin3x+2.sin2x+2.cos2x+2.sinx+4.cosx+3$
$A=(3.sinx-4.sin^3x)+4.sinx.cosx+(2-4sin^2x)+2.sinx+4.cosx+3$
$A=(5.sinx+5)+(-4.sin^3x-4sin^2x)+(4.sinx.cosx+4.cosx)$
$A=5.(sinx+1)-4.sin^2x(sinx+1)+4.cosx.(sinx+1)$
$A=(sinx+1).[5-4.(1-cos^2x)+4.cosx]$
$A=(sinx+1).(4cos^2x+4.cosx+1)$
$A=(sinx+1).(2.cosx+1)^2$

$cos2x+5=2.(2-cosx)(sinx-cosx)$

$\Leftrightarrow 2.cos^2 x -1 +5=2.(2.sinx-2.cosx-cosx.sinx+cos^2 x)$

$\Leftrightarrow cos^2 x +2=2.sinx-2.cosx-cosx.sinx+cos^2 x$

$\Leftrightarrow 2.(sinx-cosx)-cosx.sinx=2$

Đặt   $t=sinx-cosx$   ,   khi đó ta có     $\dfrac{t^2-1}{2}=(-cosx.sinx)$

pt $\Leftrightarrow 2.t+\dfrac{t^2-1}{2}=2$

.......v.........v..............

$sin(x+24^{0})+cos(x+144^{0})=cos20^{0}$

 
Đặt       $a=x+84^{0}$

 

pt $\Leftrightarrow sin\left ( a-60^{0} \right )+cos\left ( a+60^{0} \right ) = cos20^{0}$
$\Leftrightarrow sina\cdot \dfrac{1}{2}-cosa\cdot \dfrac{\sqrt{3}}{2}+cosa\cdot \dfrac{1}{2}-sina\cdot \dfrac{\sqrt{3}}{2}= cos20^{0}$
$\Leftrightarrow \dfrac{1}{2}\left ( sina+cosa \right )-\dfrac{\sqrt{3}}{2}\left ( sina+cosa \right )= cos20^{0}$
$\Leftrightarrow sin\left ( a+ 45^{0}\right )\cdot \left (\dfrac{1-\sqrt{3}}{\sqrt{2}} \right )= cos20^{0}$
$\Leftrightarrow sin\left ( x+ 129^{0}\right )= cos20^{0}:\left (\dfrac{1-\sqrt{3}}{\sqrt{2}} \right )$

 

Ko bik quá trình biến đổi có sai sót gì ko nữa ....

Lúc bấm máy tính thấy :     VP  =   -1,815346...  <  -1

Mình chỉ làm đc đến đây thôi .....

Bước đầu.... phần biến đổi có cách khác :

$2 \left ( tanx-sinx-\dfrac{1}{cosx} \right ) \ - \ 3 \left ( cotx-cosx-\dfrac{1}{sinx} \right ) \ = \ 1$
 

$\Leftrightarrow 2.tanx. \left ( 1-cosx-\dfrac{1}{sinx} \right ) \ - \ 3.cotx. \left ( 1-sinx-\dfrac{1}{cosx} \right ) \ = \ 1$
 

ĐK :    $\left\{\begin{matrix}
sinx \neq 0 \\
cosx \neq 0
\end{matrix}\right.$

 

Nhân 2 vế của phương trình cho $\left ( sinx.cosx \right ) \neq 0$ , ta có :

 

pt $\Leftrightarrow 2.\left ( sinx.cosx.tanx \right ) \ (....) - 3.\left ( cotx.sinx.cosx \right ) \ (....) \ = sinx.cosx$

 

$\Leftrightarrow 2. sin^2 x \ .\left ( 1-cosx-\dfrac{1}{sinx} \right ) \ - \ 3. cos^2 x \ .\left ( 1-sinx-\dfrac{1}{cosx} \right )=sinx.cosx$

$\Leftrightarrow 2sinx\left(sinx-sinx.cosx-1\right)+2sinxcosx=3cosx\left(cosx-sinx.cosx-1\right)+3sinxcosx$

$\Leftrightarrow 2.sinx.\left(sinx-sinx.cosx-1+cosx\right)=3.cosx.\left(cosx-sinx.cosx-1+sinx\right)$

$\Leftrightarrow 2.sinx.\left ( sinx-1 \right )\left ( 1-cosx \right )=3.cosx.\left ( 1-sinx \right )\left ( cosx-1 \right )$

.......v.......v......

Phần giải tiếp theo .... giống cách trên !!

$sin2x+cos2x+3.sinx-cosx-2=0$

pt $\Leftrightarrow 2.sinx.cosx+2.cos^2 x -1 +3.sinx-cosx-2=0$

$\Leftrightarrow \left ( 2.sinx.cosx+3.sinx \right ) + \left ( 2.cos^2 x -cosx-3 \right )=0$

Giải phương trình:

$tan^2{2x}.tan^2{3x}.tan^2{5x}=tan^2{2x}-tan^2{3x}-tan^2{5x}$

 

Câu này mình đã giải !! Bạn có thể click đg` link xem tham khảo !!

 

http://diendantoanho...2x-tan23xtan5x/

$\sum_{n=1}^{\infty }(\sqrt[n]{4}-\sqrt{\frac{n-2}{n}})$

 
Đề bài là $\sqrt{\dfrac{n-2}{n}}$ đúng ko bạn ?? Hay là   $\sqrt[n]{\dfrac{n-2}{n}}$

$2 \left ( tanx-sinx-\dfrac{1}{cosx} \right ) \ - \ 3 \left ( cotx-cosx-\dfrac{1}{sinx} \right ) \ = \ 1$

ĐK :   $\left\{\begin{matrix}
sinx \neq 0 \\
cosx \neq 0
\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
sinx \neq \ \pm \ 1 \\
cosx \neq \ \pm \ 1
\end{matrix}\right.$

pt   $\Leftrightarrow 2 \left ( \dfrac{sinx}{cosx} -sinx-\dfrac{1}{cosx} \right ) \ - \ 3 \left ( \dfrac{cosx}{sinx} -cosx-\dfrac{1}{sinx} \right ) \ = \ 3 \ - \ 2$

$\Leftrightarrow 2 \left ( \dfrac{sinx-1}{cosx} - sinx \right ) \ + \ 2 \ = \ 3 \left ( \dfrac{cosx-1}{sinx} - cosx \right ) \ + \ 3$

$\Leftrightarrow 2 \left ( \dfrac{sinx-1}{cosx} + 1 - sinx \right ) \ = \ 3 \left ( \dfrac{cosx-1}{sinx} + 1 - cosx \right )$

$\Leftrightarrow 2 \ \left ( sinx-1 \right )\left ( \dfrac{1}{cosx} - 1 \right ) \ = \ 3 \ \left ( cosx-1 \right ) \left ( \dfrac{1}{sinx} - 1 \right )$

$\Leftrightarrow \dfrac{2}{cosx} \ \left ( sinx-1 \right )\left ( 1-cosx \right ) \ = \ \dfrac{3}{sinx} \ \left ( cosx-1 \right ) \left ( 1-sinx \right )$

$\Leftrightarrow \dfrac{sinx}{cosx} \cdot \left ( sinx-1 \right )\left ( 1-cosx \right ) \ = \dfrac{3}{2} \cdot \left ( 1-cosx \right ) \left ( sinx-1 \right )$

$\Leftrightarrow \left[ \begin{array}{l} sinx = 1 \rightarrow loai. \\ cosx = 1 \rightarrow loai. \\ tanx = \dfrac{3}{2} \end{array} \right.$

 

$\Leftrightarrow x = arctan\dfrac{3}{2} + k \pi$   ,   $\left ( k \in \mathbb{Z} \right )$

Tính tổng các nghiệm nằm trong khoảng $(0;2\pi )$ của phương trình:
$(\sqrt{3}-1)sinx+(\sqrt{3}+1)cosx=2\sqrt{2}sin2x$

 

pt $\Leftrightarrow \left (\dfrac{\sqrt{3}-1}{2\sqrt{2}} \right ) \cdot \ sinx \ + \ \left (\dfrac{\sqrt{3}+1}{2\sqrt{2}} \right ) \cdot \ cosx \ = \ sin2x$

$\Leftrightarrow cos\dfrac{5 \pi}{12} \cdot \ sinx \ + \ sin\dfrac{5 \pi}{12} \cdot \ cosx \ = \ sin2x$

$\Leftrightarrow sin\left (x + \dfrac{5 \pi}{12} \right ) = sin2x$

$\Leftrightarrow \left[ \begin{array}{l} 2x = x + \dfrac{5 \pi}{12} + k2 \pi \\ 2x = \pi - x - \dfrac{5 \pi}{12} + k2 \pi \end{array} \right.$ $\left ( k \in \mathbb{Z} \right )$

$\Leftrightarrow \left[ \begin{array}{l} x = \pi \left ( \dfrac{5}{12} + 2k \right ) \\ x = \pi \left ( \dfrac{7}{36} + \dfrac{2k}{3} \right ) \end{array} \right.$ $\left ( k \in \mathbb{Z} \right )$

Vi` điều kiện   $x \in \left ( 0, 2 \pi \right )$   ,  nên phương trình đã cho có nghiệm :

 

$x \in \left \{ \dfrac{5 \pi}{12} \ ; \ \dfrac{5 \pi}{36} \ ; \ \dfrac{31 \pi}{36} \ ; \ \dfrac{55 \pi}{36} \right \}$

Do đó ta có tổng của các nghiệm là

 

$S \ = \ \dfrac{5 \pi}{12} \ + \ \dfrac{5 \pi}{36} \ + \ \dfrac{31 \pi}{36} \ + \ \dfrac{55 \pi}{36} \ = \ \dfrac{53 \pi}{18} $

pt $\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) + sin\left ( -2x+\dfrac{\pi }{6} \right ) - \ \dfrac{1}{2} = 0$

$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) + cos\left ( 2x+\dfrac{\pi }{3} \right ) - \ cos\dfrac{\pi}{3} = 0$

$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) - 2 . \ sin\left ( x+\dfrac{\pi }{3} \right ) \ . sinx = 0$

$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) \ . \left (1 - sinx \right ) = 0$

$\Leftrightarrow \left[ \begin{array}{l} sin\left ( x+\dfrac{\pi }{3} \right ) = 0 \\ sinx = 1 \end{array} \right.$

.........v...........v.............

$2.sin4x + \sqrt{3} = 3.sin2x + \sqrt{3}.cos2x$

 

$\Leftrightarrow sin2x .\left ( 4.cos2x - 3 \right ) + \sqrt{3}. \left ( 1 - cos2x \right ) = 0$

 

$\Leftrightarrow 2.sinx . cosx . \left ( 4.cos2x - 3 \right ) + \sqrt{3}. \ 2 . \ sin^2x = 0$

 

$\Leftrightarrow \left[ \begin{array}{l} 2 . sinx = 0 \\ 2 \ . 2 \ cos2x \ cosx - 3. \ cosx + \sqrt{3}. sinx = 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = k \pi \ , \ \left ( k \in \mathbb{Z} \right ) \\ 2 \ . \left ( cos3x + cosx \right ) - 3. \ cosx + \sqrt{3}. sinx = 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = k \pi \ , \ \left ( k \in \mathbb{Z} \right ) \\ cos3x = \dfrac{1}{2} \ cosx \ - \ \dfrac{\sqrt{3}}{2} \ sinx \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = k \pi \ , \ \left ( k \in \mathbb{Z} \right ) \\ cos3x = cos\left ( \dfrac{\pi}{3} +x\right ) \end{array} \right.$

........v...........v..............

$tan^2\left (x+\dfrac{\pi }{2} \right ) + cot \ x + 4 \ . cos^2\left ( x+\dfrac{\pi }{4} \right ) = 0$

$\Leftrightarrow cot^2 \ x \ + \ cot \ x \ + \ 2 \ . \left [ 1 \ + \ cos \left(2x + \dfrac{\pi }{2} \right) \right ] = 0$

ĐK :         $sin \ x \neq 0 \Leftrightarrow x \neq k \pi$  ,  $\left ( k \in \mathbb{Z} \right )$

pt $\Leftrightarrow \dfrac{cos^2 \ x}{sin^2 \ x} \ + \ \dfrac{cos \ x}{sin \ x} \ + \ 2 \ . \left ( 1 - sin \ 2x \right) = 0$

$\Leftrightarrow cos^2 \ x \ + \ cos \ x \ . sin \ x \ + \ 2 \ . sin^2 \ x \ . \left ( 1 - sin \ 2x \right) = 0$

$\Leftrightarrow \dfrac{1 + cos \ 2x}{2} \ + \dfrac{sin \ 2x}{2} \ + \ \left ( 1 - cos \ 2x \right ) . \left ( 1 - sin \ 2x \right) = 0$

$\Leftrightarrow \left (sin \ 2x + cos \ 2x \right ) \ - \ 2 \ . sin \ 2x \ cos \ 2x \ - \ 3 \ = \ 0 $

Đặt     $ \ t = sin \ 2x + cos \ 2x$  .           Khi đó    $ \ \left ( t^2-1 \right ) = 2 \ . sin \ 2x \ cos \ 2x$

 

........v.........v................

Tìm các nghiệm x thuộc $\left(0 ; 2 \pi\right)$ của phương trình
$\dfrac{sin3x - sinx}{\sqrt{1 - cos2x}} = sin2x + cos2x$

$\Leftrightarrow \dfrac{2 \ . \ cos2x \ . \ sinx}{\sqrt{2 . sin^2x}} = \sqrt{2} \ . \ cos \left ( 2x - \dfrac{\pi}{4} \right )$

$\Leftrightarrow \dfrac{cos2x  .  sinx}{\left | sinx \right |} = cos \left ( 2x - \dfrac{\pi}{4} \right )$

ĐK : __ $\left\{\begin{matrix}
sinx \neq 0 \\
x \in \left ( 0 ; 2 \pi \right )
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
sinx > 0 \\
x \in \left ( 0 ; 2 \pi \right ) \setminus \left \{ \pi \right \}
\end{matrix}\right.$

 

Dưới đk trên, ta có pt $\Leftrightarrow cos2x = cos \left ( 2x - \dfrac{\pi}{4} \right )$

........v.........v...................

Giải pt: $sin^2 \ 2x - sin^2 \ 3x = 0$

$\Leftrightarrow sin^2 \ 3x = sin^2 \ 2x $

$\Leftrightarrow \dfrac{1 - cos \ 6x }{2} = \dfrac{1 - cos \ 4x }{2}$

$\Leftrightarrow cos \ 6x = cos \ 4x $

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

$\Leftrightarrow \left[ \begin{array}{l} x = k \pi \\ x = \dfrac{\pi}{10} + \dfrac{k \pi}{5} \end{array} \right.$ $\left ( k \in \mathbb{Z} \right )$

$2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x \right)\sin x=1$

$\Leftrightarrow 2 . sinx . cos \left(\dfrac{5\pi}{12} - x \right) - \dfrac{1}{\sqrt{2}} = 0$

$\Leftrightarrow sin\dfrac{5\pi}{12} - sin\left(\dfrac{5\pi}{12} - 2x \right) - sin\dfrac{\pi}{4} = 0$

$\Leftrightarrow 2.cos\dfrac{\pi}{3}sin\dfrac{\pi}{12} = sin\left(\dfrac{5\pi}{12} - 2x \right) $

$\Leftrightarrow sin\dfrac{\pi}{12} = sin\left(\dfrac{5\pi}{12} - 2x \right) $

...........v...........v................

$sin^2x +cosx.sin^2x=cos^2x+sinx.cos^2x$

 

$\Leftrightarrow sin^2x - cos^2x + cosx.sin^2x - sinx.cos^2x = 0 $

$\Leftrightarrow \left ( sinx - cosx\right ).\left ( sinx + cosx \right ) + sinx.cosx.\left ( sinx - cosx \right ) = 0$

$\Leftrightarrow \left[ \begin{array}{l} sinx-cosx=0\\ sinx + cosx + sinx.cosx = 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} \sqrt{2}  . sin \left(x - \dfrac{\pi}{4} \right) = 0 \\ t + \dfrac{t^2-1}{2} = 0 \end{array} \right.$ __(Đặt_ $t=sinx + cosx$ __nên ta có__ $sinx.cosx = \dfrac{t^2-1}{2}$

........v......v..................

Ta có___ $ tan5x = \dfrac{tan2x + tan3x}{1-tan2x.tan3x}$

Do đó__ $ tan5x \ . \ \left ( 1 - tan2x.tan3x \right ) = tan2x + tan3x$
________________________________

$\boxed{tan^{2}2x \ . \ tan^{2}3x \ . \ tan5x = tan^{2}2x-tan^{2}3x+tan5x}$

ĐKXĐ : $\left\{\begin{matrix}
tan2x.tan3x \neq 1 \\
cos2x \neq 0 \\
cos3x \neq 0 \\
cos5x \neq 0
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
cosx \neq 0 \\
cos2x \neq 0 \\
cos3x \neq 0 \\
cos5x \neq 0
\end{matrix}\right.$

pt $\Leftrightarrow tan^{2}3x-tan^{2}2x = tan5x \ . \ \left ( 1 - tan^{2}2x \ . \ tan^{2}3x \right )$

$\Leftrightarrow tan^{2}3x-tan^{2}2x = tan5x \ . \ \left ( 1 - tan2x.tan3x \right ) \ . \ \left ( 1+tan2x.tan3x \right )$

$\Leftrightarrow \left (tan3x+tan2x \right ) \ . \ \left (tan3x-tan2x \right ) = \left (tan2x+tan3x \right ) \ . \ \left ( 1+tan2x.tan3x \right )$
 
$\Leftrightarrow \left[ \begin{array}{l} tan2x+tan3x=0\\ tan3x-tan2x = 1+tan2x.tan3x \\ \end{array} \right.$
 
$\Leftrightarrow \left[ \begin{array}{l} tan3x= - tan2x \\ \dfrac{tan3x-tan2x}{1+tan2x.tan3x} = 1 \\ \end{array} \right.$

(Vi` :__ $1+tan2x.tan3x = \dfrac{cos2x.cos3x+sin2x.sin3x}{cos2x.cos3x}= \dfrac{cos5x}{cos2x.cos3x} \neq 0 $ )

 
$\Leftrightarrow \left[ \begin{array}{l} tan3x=tan \left(-2x \right) \\ tanx = tan \dfrac{\pi}{4} \\ \end{array} \right.$
 
.........v.........v...................

$4\cos^4x-cos2x-\frac{1}{2}cos 4x+cos \frac{3x}{4}=\frac{7}{2}$

$\Leftrightarrow \left ( 2\cos^2x \right )^2 - cos2x - \dfrac{cos 4x}{2} + cos \dfrac{3x}{4} = \dfrac{7}{2}$

$\Leftrightarrow \left ( 1+cos2x \right )^2 - cos2x - \dfrac{2cos^2 2x -1}{2} + cos \dfrac{3x}{4} = \dfrac{7}{2}$

$\Leftrightarrow 1 + 2.cos2x + cos^2 2x - cos2x - cos^2 2x + \dfrac{1}{2} + cos \dfrac{3x}{4} = \dfrac{7}{2}$

$\Leftrightarrow cos2x + cos \dfrac{3x}{4} = 2$ ___(**)

Ta có__ $\left\{\begin{matrix}
cos2x \leq 1\\
cos \dfrac{3x}{4} \leq 1
\end{matrix}\right.$ _,_ $\forall x \in \mathbb{R}$
Do đó__ $ cos2x + cos \dfrac{3x}{4} \leq 2$ _,_ $\forall x \in \mathbb{R}$

Dấu "=" của pt (**) xảy ra khi và chỉ khi :

$\Leftrightarrow \left\{\begin{matrix}
cos2x = 1\\
cos \dfrac{3x}{4} = 1
\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x = k \pi\\
x = \dfrac{8k' \pi}{3}
\end{matrix}\right.$ _,_ $\left ( k , k' \in \mathbb{R} \right )$

Giao 2 họ nghiệm trên, ta có nghiệm của pt đã cho là _ $x = 2m \pi$ _,_ $\left ( m \in \mathbb{R} \right )$

$3sin^2x.cos(x+\frac{2r}{3})+2sin^3x.cosx=sinx.cos^2x+sin^2(x+\frac{r}{4}).cosx$

Đề có nhiều lỗi sai !

$3.sin^2x.cos\left ( x+\frac{2 \pi}{3} \right )+3.sin^3x.cosx=sinx.cos^2x+sin^2\left ( x+\frac{\pi}{4} \right ).cosx$

Ta có : $cos2x = \left ( cosx-sinx \right ).\left ( cosx+sinx \right )$
_______________

$\dfrac{sinx+cosx}{sinx-cosx}+2.tan2x+sin\left(\dfrac{13\pi +4x}{2} \right)=0$

ĐK : $\left\{\begin{matrix}
sinx-cosx\neq 0\\
cos2x\neq 0
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
cos2x\neq 0\\
sin2x\neq \pm 1 \end{matrix}\right.$

pt $\Leftrightarrow \dfrac{sin^2x-cos^2x}{\left ( cosx-sinx \right )^2} + sin\left(6 \pi +\dfrac{\pi}{2}+2x \right) + \dfrac{2.sin2x}{cos2x} =0$

$\Leftrightarrow \dfrac{- cos2x}{1-sin2x} + cos2x + \dfrac{2.sin2x}{cos2x} =0$

$\Leftrightarrow cos2x \left(\dfrac{1}{sin2x-1} + 1 \right) + \dfrac{2.sin2x}{cos2x} =0$

$\Leftrightarrow cos2x \left(\dfrac{sin2x}{sin2x-1}\right) + \dfrac{2.sin2x}{cos2x} =0$

$\Leftrightarrow \left[ \begin{array}{l} sin2x=0\\ \dfrac{cos2x}{sin2x-1} + \dfrac{2}{cos2x}= 0\\ \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = \dfrac{k \pi}{2} ;\left ( k\in \mathbb{Z} \right ) \\ cos^22x+2.sin2x-2= 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = \dfrac{k \pi}{2} ;\left ( k\in \mathbb{Z} \right ) \\ 1-sin^22x+2.sin2x-2= 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = \dfrac{k \pi}{2} ;\left ( k\in \mathbb{Z} \right ) \\ \left(sin2x-1 \right)^2= 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = \dfrac{k \pi}{2} ;\left ( k\in \mathbb{Z} \right ) \\ sin2x = 1 \rightarrow (Loai.) \end{array} \right.$

Vậy $x = \dfrac{k \pi}{2} ;\left ( k\in \mathbb{Z} \right )$

______________________________________________

...... giải chi tiết giúp bạn !