Bài 4: Ôn tập chương Hàm số lượng giác và phương trình lượng giác

1: \(\Leftrightarrow sinx\cdot\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\cdot cosx=0\)

=>\(sin\left(x-\dfrac{pi}{3}\right)=0\)

=>x-pi/3=kpi

=>x=kpi+pi/3

3: =>sin 3x*căn 3/2+1/2*cos3x=-1/2

=>sin(3x+pi/6)=-1/2

=>3x+pi/6=-pi/6+k2pi hoặc 3x+pi/6=7/6pi+k2pi

=>3x=-1/3pi+k2pi hoặc 3x=pi+k2pi

=>x=-1/9pi+k2pi/3 hoặc x=pi/3+k2pi/3

2.

\(sinx+cosx=\sqrt{2}sin5x\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin5x\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

1: =>2x=pi/6+k2pi hoặc 2x=5/6pi+k2pi

=>x=pi/12+kpi hoặc x=5/12pi+kpi

2: =>\(\sqrt{2}sin\left(x-\dfrac{pi}{4}\right)=0\)

=>x-pi/4=kpi

=>x=kpi+pi/4

5: =>2x=pi/4+kpi

=>x=pi/8+kpi/2

6: =>3x=2pi-pi/2+kpi

=>x=1/2pi+kpi/3

a.

\(sin^6x+cos^6x+\dfrac{3}{4}=sin2x\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\dfrac{3}{4}=sin2x\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\dfrac{3}{4}=sin2x\)

\(\Leftrightarrow\dfrac{7}{4}-\dfrac{3}{4}sin^22x=sin2x\)

\(\Leftrightarrow3sin^22x+4sin2x-7=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\dfrac{7}{3}< -1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow2x=\dfrac{\pi}{2}+k2\pi\)

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

b.

ĐKXĐ; \(x\ne\dfrac{k\pi}{2}\)

\(2tanx+cotx=2sin2x+\dfrac{1}{sin2x}\)

\(\Leftrightarrow\dfrac{2sinx}{cosx}+\dfrac{cosx}{sinx}=2sin2x+\dfrac{1}{2sinx.cosx}\)

\(\Rightarrow4sin^2x+2cos^2x=2sin^22x+1\)

\(\Leftrightarrow2sin^2x+2=2sin^22x+1\)

\(\Leftrightarrow2-cos2x=2\left(1-cos^22x\right)\)

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

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

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

c.

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(\dfrac{sin^22x-2}{sin^22x-4cos^2x}=tan^2x\)

\(\Leftrightarrow\dfrac{sin^22x-2}{4sin^2x.cos^2x-4cos^2x}=\dfrac{sin^2x}{cos^2x}\)

\(\Leftrightarrow\dfrac{sin^22x-2}{4cos^2x\left(sin^2x-1\right)}=\dfrac{sin^2x}{cos^2x}\)

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

\(\Rightarrow2-sin^22x=4sin^2x.cos^2x\)

\(\Leftrightarrow2-sin^22x=sin^22x\)

\(\Leftrightarrow1-sin^22x=0\)

\(\Leftrightarrow cos^22x=0\)

\(\Leftrightarrow cos2x=0\)

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

a.

ĐKXĐ: \(x\ne k\pi\)

\(\dfrac{3}{sin^2x}+\left(6+\sqrt{3}\right)cotx+2\sqrt{3}-3=0\)

\(\Leftrightarrow3\left(1+cot^2x\right)+\left(6+\sqrt{3}\right)cotx+2\sqrt{3}-3=0\)

\(\Leftrightarrow3cot^2x+\left(6+\sqrt{3}\right)cotx+2\sqrt{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cotx=-2\\cotx=-\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

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

b.

\(4sin^2x+2\left(\sqrt{3}+1\right)cosx-\sqrt{3}=4\)

\(\Leftrightarrow4\left(1-cos^2x\right)+2\left(\sqrt{3}+1\right)cosx-\sqrt{3}=4\)

\(\Leftrightarrow-4cos^2x+2\left(\sqrt{3}+1\right)cosx-\sqrt{3}=0\)

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

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

46.

\(cos2x+\left(2m+1\right)cosx+2m=0\)

\(\Leftrightarrow2cos^2x-1+\left(2m+1\right)cosx+2m=0\)

\(\Leftrightarrow2cos^2x+cosx-1+2m\left(cosx+1\right)=0\)

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

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

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

Do (1) có đúng 1 nghiệm \(x=\pi\) thuộc khoảng đã cho nên pt đã cho có nghiệm duy nhất thuộc \([0;2\pi)\) khi và chỉ khi:

\(\left[{}\begin{matrix}\dfrac{2m-1}{2}=-1\\\dfrac{2m-1}{2}>1\\\dfrac{2m-1}{2}< -1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m>\dfrac{3}{2}\\m\le-\dfrac{1}{2}\end{matrix}\right.\)

47.

\(\sqrt{3}sinx+cosx=m\)

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

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=\dfrac{m}{2}\)

Đặt \(x+\dfrac{\pi}{6}=t\Rightarrow t\in\left[0;3\pi\right]\)

Từ đường tròn lượng giác ta thấy \(sint=\dfrac{m}{2}\) trên \(\left[0;3\pi\right]\):

- Có 1 nghiệm khi \(\dfrac{m}{2}=-1\)

- Có 2 nghiệm khi \(-1< \dfrac{m}{2}< 0\) hoặc \(\dfrac{m}{2}=1\)

- Có 4 nghiệm khi \(0\le\dfrac{m}{2}< 1\)

\(\Rightarrow\) Không tồn tại m để pt đã cho có đúng 3 nghiệm trên miền đã cho 

Cả 4 đáp án đều sai

48.

\(sin5x=\left(m-2\right)sinx\)

\(\Leftrightarrow sin5x-sinx=\left(m-3\right)sinx\)

\(\Leftrightarrow2cos3x.sin2x-\left(m-3\right)sinx=0\)

\(\Leftrightarrow4cos3x.cosx.sinx-\left(m-3\right)sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(1\right)\\4cos3x.cosx-m+3=0\left(2\right)\end{matrix}\right.\)

(1) có đúng 1 nghiệm \(x=0\) thuộc đoạn đã cho

Xét (2): \(\Leftrightarrow2cos4x+2cos2x-m+3=0\)

\(\Leftrightarrow4cos^22x-2+2cos2x-m+3=0\)

\(\Leftrightarrow4cos^22x+2cos2x+1=m\)

Đặt \(2x=t\in\left[-\dfrac{2\pi}{3};\dfrac{\pi}{2}\right]\)

\(\Rightarrow4cos^2t+2cost+1=m\)

Trên \(\left[-\dfrac{2\pi}{3};\dfrac{\pi}{2}\right]\), pt \(cost=k\) có:

- Đúng 1 nghiệm khi \(k=1\) hoặc \(-\dfrac{1}{2}< k< 0\)

- Có 2 nghiệm khi \(0\le k< 1\)

Do đó bài toán thỏa mãn khi: \(4k^2+2k+1=m\) có nghiệm: \(\left[{}\begin{matrix}-\dfrac{1}{2}< k_1< k_2< 0\\\left[{}\begin{matrix}0\le k_1< 1\le k_2\\k_1< 0\le k_2< 1\end{matrix}\right.\end{matrix}\right.\)

\(\left\{{}\begin{matrix}m\ne1\\\dfrac{3}{4}< m< 7\end{matrix}\right.\)

7.

\(3cos^2x=-8cosx-5\)

\(\Leftrightarrow3cos^2x+8cosx+5=0\)

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

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

8.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{6}\)

9.

\(sin^2x-4sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=4>1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=k\pi\)

10.

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

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

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

\(\Leftrightarrow3sin3x-4sin^33x-\sqrt{3}cos9x=1\)

\(\Leftrightarrow sin9x-\sqrt{3}cos9x=1\)

\(\Leftrightarrow\dfrac{1}{2}sin9x-\dfrac{\sqrt{3}}{2}cos9x=\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(9x-\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow...\)