Bài 3: Một số phương trình lượng giác thường gặp

ĐK: \(x\ne\dfrac{\pi}{4}+k\pi;x\ne\dfrac{k\pi}{2}\)

\(\dfrac{2sin^2x+cos4x-cos2x}{\left(sinx-cosx\right)sin2x}=0\)

\(\Leftrightarrow2sin^2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sin^2x-1+cos4x-cos2x+1=0\)

\(\Leftrightarrow2cos^22x-2cos2x=0\)

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

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

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

Đối chiếu điều kiện ta được \(x=-\dfrac{\pi}{4}+k\pi\)

1.

a, Phương trình có nghiệm khi: 

\(\left(m+2\right)^2+m^2\ge4\)

\(\Leftrightarrow m^2+4m+4+m^2\ge4\)

\(\Leftrightarrow2m^2+4m\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}m\ge0\\m\le-2\end{matrix}\right.\)

b, Phương trình có nghiệm khi:

\(m^2+\left(m-1\right)^2\ge\left(2m+1\right)^2\)

\(\Leftrightarrow2m^2+6m\le0\)

\(\Leftrightarrow-3\le m\le0\)

2.

a, Phương trình vô nghiệm khi:

\(\left(2m-1\right)^2+\left(m-1\right)^2< \left(m-3\right)^2\)

\(\Leftrightarrow4m^2-4m+1+m^2-2m+1< m^2-6m+9\)

\(\Leftrightarrow4m^2-7< 0\)

\(\Leftrightarrow-\dfrac{\sqrt{7}}{2}< m< \dfrac{\sqrt{7}}{2}\)

b, \(2sinx+cosx=m\left(sinx-2cosx+3\right)\)

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

 Phương trình vô nghiệm khi:

\(\left(m-2\right)^2+\left(2m+1\right)^2< 9m^2\)

\(\Leftrightarrow m^2-4m+4+4m^2+4m+1< 9m^2\)

\(\Leftrightarrow m^2-1>0\)

\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)

1.

c, \(\left(m+2\right)sin2x+mcos^2x=m-2+msin^2x\)

\(\Leftrightarrow\left(m+2\right)sin2x+m\left(cos^2x-sin^2x\right)=m-2\)

\(\Leftrightarrow\left(m+2\right)sin2x+mcos2x=m-2\)

Phương trình vô nghiệm khi:

\(\left(m+2\right)^2+m^2< \left(m-2\right)^2\)

\(\Leftrightarrow m^2+4m+4+m^2< m^2-4m+4\)

\(\Leftrightarrow m^2+8m< 0\)

\(\Leftrightarrow-8\le m\le0\)

1.

\(\left(sin2x+\sqrt{3}cos2x\right)^2=2cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left(\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x\right)^2=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow sin^2\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow cos^2\left(\dfrac{\pi}{2}-2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow cos^2\left(2x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{6}\right)\left[cos\left(2x-\dfrac{\pi}{6}\right)-\dfrac{1}{2}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\cos\left(2x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)

2.

ĐK: \(x\ne\dfrac{k\pi}{2};x\ne\pi+k2\pi\)

\(cotx+\left(1+tanx.tan\dfrac{x}{2}\right).sinx=4\)

\(\Leftrightarrow\dfrac{cosx}{sinx}+\left(1+\dfrac{sinx.sin\dfrac{x}{2}}{cosx.cos\dfrac{x}{2}}\right).sinx=4\)

\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{cosx.cos\dfrac{x}{2}+sinx.sin\dfrac{x}{2}}{cosx.cos\dfrac{x}{2}}.sinx=4\)

\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{cos\left(x-\dfrac{x}{2}\right).sinx}{cosx.cos\dfrac{x}{2}}=4\)

\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{cos\dfrac{x}{2}.sinx}{cosx.cos\dfrac{x}{2}}=4\)

\(\Leftrightarrow\dfrac{cosx}{sinx}+\dfrac{sinx}{cosx}=4\)

\(\Leftrightarrow\dfrac{cos^2x+sin^2x}{sinx.cosx}=4\)

\(\Leftrightarrow\dfrac{2}{sin2x}=4\)

\(\Leftrightarrow sin2x=\dfrac{1}{2}\)

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

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

3.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(cotx-tanx=\dfrac{2}{sin2x}-4sin2x\)

\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=\dfrac{2}{sin2x}-4sin2x\)

\(\Leftrightarrow\dfrac{2cos2x}{sin2x}=\dfrac{2-4sin^22x}{sin2x}\)

\(\Leftrightarrow\dfrac{2cos2x}{sin2x}=\dfrac{2cos2x}{sin2x}\)

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

Ta có : \(\cos\left(2x+\dfrac{\pi}{6}\right)=m+1,x\in\left(\dfrac{7\pi}{24};\dfrac{3\pi}{4}\right)\)

Thấy \(x\in\left(\dfrac{7\pi}{24};\dfrac{3\pi}{4}\right)\)

\(\Rightarrow2x+\dfrac{\pi}{6}\in\left(\dfrac{3\pi}{4};\dfrac{5\pi}{3}\right)\)

\(\Rightarrow\cos\left(2x+\dfrac{\pi}{6}\right)\in\left(-1;\dfrac{1}{2}\right)\)

\(\Rightarrow-1< m+1< \dfrac{1}{2}\)

\(\Rightarrow-2< m< -\dfrac{1}{2}\)

Vậy ...

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(4sin^2x+tanx+\sqrt{2}\left(1+tanx\right)sin3x=1\)

\(\Leftrightarrow4sin^2x-2+1+tanx+\sqrt{2}\left(1+tanx\right)sin3x=0\)

\(\Leftrightarrow-2cos2x+\left(1+tanx\right)\left(1+\sqrt{2}sin3x\right)=0\)

\(\Leftrightarrow2sin^2x-2cos^2x+\dfrac{sinx+cosx}{cosx}\left(1+\sqrt{2}sin3x\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(2sinx-2cosx+\dfrac{1+\sqrt{2}sin3x}{cosx}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\2sinx-2cosx+\dfrac{1+\sqrt{2}sin3x}{cosx}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\2sinx.cosx-2cos^2x+1+\sqrt{2}sin3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\sin2x-cos2x+\sqrt{2}sin3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)+\sqrt{2}sin3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\2sin\left(\dfrac{5x}{2}-\dfrac{\pi}{8}\right).cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\sin\left(\dfrac{5x}{2}-\dfrac{\pi}{8}\right)=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=k\pi\\\dfrac{5x}{2}-\dfrac{\pi}{8}=k\pi\\\dfrac{x}{2}+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

Pt tương đương

\(\left\{{}\begin{matrix}4sin^2x+\dfrac{sinx}{cosx}+\sqrt{2}\left(1+\dfrac{sinx}{cosx}\right).sin3x=1\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}4sin^2x.cosx+sinx+\sqrt{2}\left(sinx+cosx\right).sin3x=cosx\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}2sin2x.sinx+sinx+\sqrt{2}\left(sinx+cosx\right).sin3x=cosx\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}sinx-cos3x+\sqrt{2}\left(sinx+cosx\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}sinx-sin\left(3x+\dfrac{\pi}{2}\right)-2sin\left(x+\dfrac{\pi}{4}\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}2cos\left(2x+\dfrac{\pi}{4}\right).sin\left(-x-\dfrac{\pi}{4}\right)-2sin\left(x+\dfrac{\pi}{4}\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}2cos\left(2x+\dfrac{\pi}{4}\right).sin\left(x+\dfrac{\pi}{4}\right)+2sin\left(x+\dfrac{\pi}{4}\right).sin3x=0\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)\left[cos\left(2x+\dfrac{\pi}{4}\right)+sin3x\right]=0\\cosx\ne0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\sin3x=sin\left(2x+\dfrac{3\pi}{4}\right)\end{matrix}\right.\\x\ne\dfrac{\pi}{2}+k\pi,k\in Z\end{matrix}\right.\)

Giải nốt nhé, toàn phương trình cơ bản rồi

1.

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{2}{3}\end{matrix}\right.\)

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

2.

\(1-cos^2\dfrac{x}{2}-2cos\dfrac{x}{2}+2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{x}{2}=1\\cos\dfrac{x}{2}=-3< -1\left(loại\right)\end{matrix}\right.\)

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

\(\Leftrightarrow x=k4\pi\)

3.

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

\(\Leftrightarrow-3sin^2x-2sinx+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{5}{3}< -1\left(loại\right)\end{matrix}\right.\)

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

4.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=4cos^22x\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=4\left(1-sin^22x\right)\)

\(\Leftrightarrow sin^22x=\dfrac{12}{13}\)

\(\Leftrightarrow\dfrac{1-cos4x}{2}=\dfrac{12}{13}\)

\(\Leftrightarrow cos4x=-\dfrac{11}{13}\)

\(\Leftrightarrow x=\pm\dfrac{1}{4}arccos\left(-\dfrac{11}{13}\right)+\dfrac{k\pi}{2}\)

5.

\(1-cos^2x-\dfrac{1}{4}=cos^4x\)

\(\Leftrightarrow cos^4x+cos^2x-\dfrac{3}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{1}{2}\\cos^2x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2cos^2x=1\)

\(\Leftrightarrow2cos^2x-1=0\)

\(\Leftrightarrow cos2x=0\)

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

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

`(m^2+m)cos2x=m^2-m-3-m^2 cos2x`

`<=> (2m^2+m)cos2x=m^2-m-3`

`<=>cos2x =(m^2-m-3)/(2m^2+m)`

PT có nghiệm `<=> -1 <= (m^2-m-3)/(2m^2+m) <=1`

`<=> [(m<=-1 \vee m>=1),(-1/2 < m <0):}`