Đáp án:

`b)` \(\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{3\pi}{2}+k2\pi\end{array}\quad (k \in \Bbb Z) \right.\)

`c)` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{array}\quad (k \in \Bbb Z) \right.\)

`d)` \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=k2\pi\end{array}\quad (k \in \Bbb Z) \right.\)

Giải thích các bước giải:

`b) \quad \sinx-cosx+4sinxcosx+1=0` 

Đặt `sinx-cosx=t\ (-sqrt2<=t<=sqrt2)`

`=>(sinx-cosx)^2=t^2`

`<=>sin^2x-2sinxcosx+cos^2x=t^2`

`<=> -2sinxcosx=t^2-1`

`<=> 4sinxcosx=2-2t^2`

`PT<=>t+2-2t^2+1=0`

`<=> 2t^2-t-3=0`

`<=>`\(\left[ \begin{array}{l}t=-1\ (\text{nhận})\\t=\dfrac{3}{2}\ (\text{loại})\end{array} \right.\)

Với `t=-1=>sinx-cosx=-1`

`<=>sqrt2sin(x-pi/4)=-1`

`<=>sin(x-pi/4)=-(sqrt2)/2`

`<=>sin(x-pi/4)=sin(-pi/4)`

`<=>`\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{3\pi}{2}+k2\pi\end{array}\quad (k \in \Bbb Z) \right.\)

`c)\quad sin 2x-12(sinx-cosx)+12=0`

Đặt `sinx-cosx=t\ (-sqrt2<=t<=sqrt2)`

`=>(sinx-cosx)^2=t^2`

`<=>sin^2x-2sinxcosx+cos^2x=t^2`

`<=>2sinxcosx=1-t^2`

`<=>sin2x=1-t^2`

`PT<=>1-t^2-12t+12=0`

`<=> t^2+12t-13=0`

`<=>`\(\left[ \begin{array}{l}t=1\ (\text{nhận})\\t=-13\ (\text{loại})\end{array} \right.\) 

Với `t=1=>sinx-cosx=1`

`<=>sqrt2sin(x-pi/4)=1`

`<=>sin(x-pi/4)=(sqrt2)/2`

`<=>sin(x-pi/4)=sin\ pi/4`

`<=>`\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{array}\quad (k \in \Bbb Z) \right.\)

`d)\quad sin^3x+cos^3x=1`

`<=>(sinx+cosx)(sin^2x-sinxcosx+cos^2x)=1`

`<=>(sinx+cosx)(1-sinxcosx)=1\quad(**)`

Đặt `sinx+cosx=t\ (-sqrt2<=t<=sqrt2)`

`=>(sinx+cosx)^2=t^2`

`<=>sin^2x+2sinxcosx+cos^2x=t^2`

`<=>2sinxcosx=t^2-1`

`<=>sinxcosx=(t^2-1)/2`

`(**)<=>t.(1-(t^2-1)/2)=1`

`<=>t^3-3t+2=0`

`<=>`\(\left[ \begin{array}{l}t=1\ (\text{nhận})\\t=-2\ (\text{loại})\end{array} \right.\) 

Với `t=1=>sinx+cosx=1`

`<=>sqrt2cos(x-pi/4)=1`

`<=>cos(x-pi/4)=(sqrt2)/2`

`<=>cos(x-pi/4)=cos\ pi/4`

`<=>`\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=k2\pi\end{array}\quad (k \in \Bbb Z) \right.\)

`a,3(sinx + cosx)+2sin2x+3=0`

`<=>3(sinx + cosx)+4sinxcosx+3=0`

Đặt `sinx + cosx = t (|t|<=sqrt2)`

`=>sinx . cosx = (t^2-1)/2`

Phương trình tương đương:

`3t+4 . (t^2-1)/2 +3=0`

`<=>3t+2t^2-2+3=0`

`<=>2t^2+3t+1=0`

`<=>`\(\left[ \begin{array}{l}t=-1(tm)\\t=-\dfrac{1}{2}(tm)\end{array} \right.\) 

`TH1: t=-1`

`<=> sinx +cosx = -1`

`<=> sqrt2 . sin(x+ pi/4)=-1`

`<=>sin(x+ pi/4) = -1/sqrt2`

`=>`\(\left[ \begin{array}{l}x+\dfrac{\pi}{4}=\dfrac{-\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{-\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{-\pi}{2}+k2\pi\\x=\pi+k2\pi(k \in Z)\end{array} \right.\) 

`TH2: t =-1/2`

`<=> sqrt2 . sin(x+ pi/4)=-1/2`

`<=>  sin(x+ pi/4)=-sqrt2/4`

`<=>`\(\left[ \begin{array}{l}x+\dfrac{\pi}{4}=arcsin(-\dfrac{\sqrt{2}}{4})+h2\pi\\x+\dfrac{\pi}{4}=\pi -arcsin(-\dfrac{\sqrt{2}}{4})+h2\pi\end{array} \right.\) 

`<=>``<=>`\(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+arcsin(-\dfrac{\sqrt{2}}{4})+h2\pi\\x=\dfrac{3\pi}{4} -arcsin(-\dfrac{\sqrt{2}}{4})+h2\pi(h \in Z)\end{array} \right.\) 

`b, Đặt `sinx -cosx =t (|t| <= sqrt2)`

`=> sinx cosx = (1-t^2)/2`

`sinx - cosx +4sinx cosx +1=0`

`<=>t+ 4 . (1-t^2)/2 +1=0`

`<=>t+2-2t^2+1=0`

`<=>-2t^2 +t +3=0`

\(\left[ \begin{array}{l}t=-1(tm)\\t=\dfrac{3}{2}(ktm)\end{array} \right.\) 

`<=>sqrt2 . sin(x- pi/4) = -1`

`<=>sin(x- pi/4) = -1/sqrt2`

`<=>`\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{-\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\pi -\dfrac{-\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{3\pi}{2}+k2\pi(k \in Z)\end{array} \right.\) 

`c, sin2x - 12(sinx-cosx)+12=0`

`<=>2sinx cosx - 12(sinx-cosx)+12=0`

 Đặt `sinx -cosx =t (|t| <= sqrt2)`

`=> sinx cosx = (1-t^2)/2`

Phương trình tương đương:

`2. (1-t^2)/2 -12t+12=0`

`<=>1-t^2-12t+12=0`

`<=>-t^2-12t+13=0`

`<=>`\(\left[ \begin{array}{l}t=1(tm)\\t=-13(ktm)\end{array} \right.\) 

`<=>sqrt2 . sin(x-pi/4)=1`

`<=>sin(x-pi/4) = 1/sqrt2`

`<=>`\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\pi -\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+ k2\pi\\x=\pi+k2\pi(k \in Z)\end{array} \right.\) 

`d,sin^3x+cos^3x=1`

`<=>(sinx + cosx)^3 -3 sinx cosx (sinx + cosx)=1`

Đặt `sinx + cosx = t (|t|<=sqrt2)`

`=>sinx . cosx = (t^2-1)/2`

Phương trình tương đương:

`t^3 -3. (t^2 -1)/2 .t -1=0`

`<=>-t^3 +3t-2=0`

`<=>``<=>`\(\left[ \begin{array}{l}t=1(tm)\\t=-2(ktm)\end{array} \right.\) 

`<=>sqrt2 . sin(x+pi/4)=1`

`<=>sin(x+pi/4) = 1/sqrt2`

`<=>`\(\left[ \begin{array}{l}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi -\dfrac{\pi}{4}+k2\pi\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x= k2\pi\\x=\dfrac{\pi}{2}+k2\pi(k \in Z)\end{array} \right.\)