ĐKXĐ: \(sinx\ne0\)

\(2cos^2x-3cosx+1=sinx-2sinx^2cosx+2cos^2x.sinx\)

\(\Leftrightarrow2cos^2x\left(1-sinx\right)+1-sinx-3cosx+2sin^2x.cosx=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(2cos^2x+1\right)-cosx\left(3-2sin^2x\right)=0\)

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

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

\(\Leftrightarrow sinx+cosx=1\)

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

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

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

a) Pt\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x\left(sin^2x+cos^2x\right)+3sinx.cosx-\dfrac{m}{4}+2=0\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x-\dfrac{3}{2}sin2x-\dfrac{m}{4}+2=0\)

\(\Leftrightarrow-3sin^22x-6sin2x-m+12=0\)

Đặt \(t=sin2x;t\in\left[-1;1\right]\)

Pttt: \(-3t^2-6t-m+12=0\)

\(\Leftrightarrow-3t^2-6t+12=m\) (1)

Đặt \(f\left(t\right)=-3t^2-6t+12;t\in\left[-1;1\right]\) 

Vẽ BBT sẽ tìm được \(f\left(t\right)_{min}=3;f\left(t\right)_{max}=15\)\(\Leftrightarrow3\le f\left(t\right)\le15\)\(\Rightarrow m\in\left[3;15\right]\) thì pt (1) sẽ có nghiệm

mà \(m\in Z\) nên tổng m nguyên để pt có nghiệm là 13 m

Vậy có tổng 13 m nguyên

b) Pt\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(1\right)\\2cos^2x-\left(2m+1\right)cosx+m=0\left(2\right)\end{matrix}\right.\)

Từ (1)\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)

\(x\in\left[0;2\pi\right]\Rightarrow0\le\dfrac{\pi}{2}+k2\pi\le2\pi\)\(\Leftrightarrow-\dfrac{1}{4}\le k\le\dfrac{3}{4}\)\(\Rightarrow k=0\)

Tại k=0\(\Rightarrow x=\dfrac{\pi}{2}\)

Để pt ban đầu có 4 nghiệm pb \(\in\left[0;2\pi\right]\)

\(\Leftrightarrow\) Pt (2) có 3 nghiệm pb khác \(\dfrac{\pi}{2}\)

Xét pt (2) có: \(2cos^2x-\left(2m+1\right)cosx+m=0\)

Vì là phương trình bậc hai ẩn \(cosx\) nên pt (2) chỉ có nhiều nhất ba nghiệm \(\Leftrightarrow\) Pt (2) có một nghiệm cosx=0

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) mà \(x\ne\dfrac{\pi}{2}\)

\(\Rightarrow\) Pt (2) chỉ có nhiều nhất hai nghiệm

\(\Rightarrow\) Pt ban đầu không thể có 4 nghiệm phân biệt

Vậy \(m\in\varnothing\) 

a/

\(\Leftrightarrow3\left(cos4x+1\right)+2cos^2x\left(1-4cos^4x\right)=0\)

\(\Leftrightarrow3\left(2cos^22x-1+1\right)+2cos^2x\left(1-2cos^2x\right)\left(1+2cos^2x\right)=0\)

\(\Leftrightarrow6cos^22x+\left(1+cos2x\right).\left(-cos2x\right)\left(2+cos2x\right)=0\)

Đặt \(cos2x=a\)

\(\Rightarrow6a^2-a\left(a+1\right)\left(a+2\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\\cos2x=2\left(l\right)\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=k\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow4+3sinx+sin^3x=3\left(1-sin^2x\right)+\left(1-sin^2x\right)^3\)

Đặt \(sinx=a\) ta được:

\(a^3+3a+4=3-3a^2+\left(1-a\right)^3\)

\(\Leftrightarrow a^3+3a^2+3a+1=\left(1-a\right)^3\)

\(\Leftrightarrow\left(a+1\right)^3=\left(1-a\right)^3\)

\(\Leftrightarrow a+1=1-a\)

\(\Leftrightarrow a=0\)

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

c/

ĐKXĐ: ...

\(\Leftrightarrow2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=2cos^2x-4\)

\(\Leftrightarrow2cos^2x+2cos^2x.tanx.tan\frac{x}{2}=2cos^2x-4\)

\(\Leftrightarrow cos^2x.tanx.tan\frac{x}{2}=-2\)

\(\Leftrightarrow sinx.cosx.tan\frac{x}{2}=-2\)

\(\Leftrightarrow sinx.cosx.\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=-2\)

\(\Leftrightarrow sinx.cosx.\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}.cos\frac{x}{2}}=-1\)

\(\Leftrightarrow cosx\left(\frac{1-cosx}{2}\right)=-1\)

\(\Leftrightarrow cos^2x-cosx-2=0\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pi+k2\pi\)

a/

DKXD: ...

\(\Leftrightarrow-cos2x.tan^22x+3.cos2x=0\)

\(\Leftrightarrow cos2x\left(3-tan^22x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\tan^22x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{\pi}{6}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

DKXD: ...

\(\Leftrightarrow\frac{sinx}{cosx}+\frac{sin2x}{cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow\frac{sinx.cos2x+sin2x.cosx}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow\frac{sin\left(2x+x\right)}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)

\(\Leftrightarrow sin3x\left(\frac{1}{cosx.cos2x}-\frac{2}{sin2x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\left(1\right)\\2cosx.cos2x=sin2x\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3sinx-4sin^3x=0\) (tìm nghiệm thẳng bằng \(3x=k\pi\) rồi dựa vào đường tròn lượng giác loại nghiệm cũng được)

\(\Leftrightarrow sinx\left(3-4sin^2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\sinx=\pm\frac{\sqrt{3}}{2}\end{matrix}\right.\)

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

\(\left(2\right)\Leftrightarrow2cosx.cos2x=2sinx.cosx\)

\(\Leftrightarrow2cosx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(l\right)\\cos2x=sinx=cos\left(\frac{\pi}{2}-x\right)\end{matrix}\right.\)

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

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

c/

\(\Leftrightarrow sinx.cos2x-sinx+1-cos2x=0\)

\(\Leftrightarrow sinx\left(cos2x-1\right)-\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\cos2x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\2x=k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow2cos^2x-1+2cosx-2\cdot sinx\cdot cosx+1=0\)

=>\(2\cdot cos^2x+2cosx-2sinx\cdot cosx=0\)

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

=>\(cosx\left(sinx-cosx-1\right)=0\)

=>\(cosx\left[\sqrt{2}\cdot sin\left(x-\dfrac{pi}{4}\right)-1\right]=0\)

=>\(\left[{}\begin{matrix}cosx=0\\sin\left(x-\dfrac{pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x-\dfrac{pi}{4}=\dfrac{pi}{4}+k2pi\\x-\dfrac{pi}{4}=\dfrac{3}{4}pi+k2pi\end{matrix}\right.\)

=>x=pi/2+kpi hoặc x=pi+k2pi

1.

\(\Leftrightarrow cos\left(2x+\dfrac{4\pi}{3}\right)=0\)

\(\Leftrightarrow2x+\dfrac{4\pi}{3}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow2x=-\dfrac{5\pi}{6}+k\pi\)

\(\Leftrightarrow x=-\dfrac{5\pi}{12}+\dfrac{k\pi}{2}\)

b.

\(\Leftrightarrow2+2cos\left(2x+\dfrac{\pi}{3}\right)-3=0\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

c.

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

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

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

a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)

\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)

b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx+\sqrt{3}\cdot cosx=0\)

\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)

\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)

(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)

\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)

\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)

a.

ĐKXĐ: \(cosx\ne0\)

Chia 2 vế cho \(cos^2x\) ta được:

\(\left(1+tanx\right).tan^2x=3tanx\left(1-tanx\right)+\frac{3}{cos^2x}\)

\(\Leftrightarrow tan^2x\left(tanx+1\right)=3tanx-3tan^2x+3+3tan^2x\)

\(\Leftrightarrow tan^2x\left(tanx+1\right)-3\left(tanx+1\right)=0\)

\(\Leftrightarrow\left(tan^2x-3\right)\left(tanx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)

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

c/

\(\Leftrightarrow cos^3x=sinx\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(\Leftrightarrow cos^3x=sinx\left(cos2x-\frac{1}{2}\right)\)

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

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

\(\Leftrightarrow2cos^3x=sinx-4sin^3x\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow2=tanx\left(1+tan^2x\right)-4tan^3x\)

\(\Leftrightarrow3tan^3x-tanx+2=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(3tan^2x-3tanx+2\right)=0\)

\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)

d/

\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(sinx+cosx\right)-4cos^3x\left(sin^2x+cos^2x+2sinx.cosx\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)^2-4cos^3x\left(sinx+cosx\right)^2=0\)

\(\Leftrightarrow\left(cosx-sinx-4cos^3x\right)\left(sinx+cosx\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx-4cos^3x=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=k\pi\)

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

Xét \(\left(2\right)\), nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)

\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)

\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)

\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)