Bài 2: Phương trình lượng giác cơ bản - Hỏi đáp

Đặt \(tan\frac{x}{2}=t\Rightarrow t\ne0\)

\(\Rightarrow\left(2m-1\right)t+m=0\) có nghiệm \(t\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\2m-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\m\ne\frac{1}{2}\end{matrix}\right.\)

\(sin2x=sin\left(\frac{\pi}{3}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}-x+k2\pi\\2x=\pi-\frac{\pi}{3}+x+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Bạn không cho biết tính tổng các nghiệm trong khoảng nào, đoạn nào thì không thể tính được, vì pt lượng giác có vô số nghiệm

2/ ĐKXĐ: \(cos3x\ne1\Rightarrow3x\ne k2\pi\Rightarrow x\ne\frac{k2\pi}{3}\)

Phương trình tương đương:

\(sin2x=0\Leftrightarrow2x=k\pi\Rightarrow x=\frac{k\pi}{2}\)

Kết hợp ĐKXĐ \(\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(\Rightarrow\) nghiệm dương nhỏ nhất là \(x=\frac{\pi}{2}\) khi \(k=0\)

Lời giải:

$\sin (-2x+3)=\frac{-\sqrt{3}}{2}$

\(\Rightarrow \left[\begin{matrix} -2x+3=\frac{-\pi}{3}+2k\pi \\ -2x+3=\frac{4}{3}\pi +2k\pi \end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{\pi}{6}+\frac{3}{2}-k\pi\\ x=\frac{-2}{3}\pi +\frac{3}{2}-k\pi \end{matrix}\right.\) với $k$ nguyên.

Phương trình lượng giác bậc nhất cơ bản mà :(

\(\Leftrightarrow\sin x-\cos x=\frac{\sqrt{2}}{2}\Leftrightarrow\sin\left(x-\frac{\pi}{4}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{12}+k2\pi\\x=\frac{13}{12}\pi+k2\pi\end{matrix}\right.\)

\(th1:0\le\frac{5\pi}{12}+k2\pi\le2\pi\)

\(th2:0\le\frac{13}{12}\pi+k2\pi\le2\pi\)

Chặn k là okie :)

\(\Leftrightarrow x+10^0=-90^0+k360^0\)

\(\Leftrightarrow x=-100^0+k360^0\)

ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\)

\(\frac{1}{\frac{sinx}{cosx}+\frac{cos2x}{sin2x}}=\frac{\sqrt{2}\left(cosx-sinx\right)}{\frac{cosx}{sinx}-1}\)

\(\Leftrightarrow\frac{sin2x.cosx}{cos2x.cosx+sin2x.sinx}=\frac{\sqrt{2}sinx\left(cosx-sinx\right)}{cosx-sinx}\)

\(\Leftrightarrow\frac{sin2x.cosx}{cosx}=\sqrt{2}sinx\)

\(\Leftrightarrow2sinx.cosx=\sqrt{2}sinx\)

\(\Leftrightarrow cosx=\frac{\sqrt{2}}{2}\)

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

Vậy \(x=-\frac{\pi}{4}+k2\pi\)

d. ĐKXĐ: ...

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)cot\left(\frac{\pi}{2}-\left(x+\frac{\pi}{3}\right)\right)\)

\(\Leftrightarrow1-\frac{1}{2}sin^22x=\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)tan\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow1-\frac{1}{4}\left(1-cos4x\right)=\frac{7}{8}\)

\(\Leftrightarrow cos4x=\frac{1}{2}\)

\(\Leftrightarrow x=...\)

c/

ĐKXĐ: ....

\(\frac{1}{sinx}+\frac{1}{sin\left(x+\frac{\pi}{2}-2\pi\right)}=4sin\left(2\pi-\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow\frac{1}{sinx}+\frac{1}{cosx}=-4sin\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow\frac{sinx+cosx}{sinx.cosx}=-4sin\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow\frac{\sqrt{2}sin\left(x+\frac{\pi}{4}\right)}{sinx.cosx}+4sin\left(x+\frac{\pi}{4}\right)=0\)

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

\(\left(1\right)\Leftrightarrow\sqrt{2}+4sinx.cosx=0\)

\(\Leftrightarrow sin2x=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow x=...\)

4.

ĐKXĐ: \(2cos^2x+sinx-1\ne0\)

\(\Leftrightarrow-2sin^2x+sinx+1\ne0\Rightarrow\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)

Khi đó pt tương đương:

\(\Leftrightarrow\frac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)

\(\Leftrightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\left(loại\right)\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

3.

\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)

Bạn thử đưa ra 1 bài toán cụ thể làm ví dụ xem. Mình chưa hiểu ý bạn nói lắm @@

d.

\(\Leftrightarrow cos3x-cosx+cos2x-1=0\)

\(\Leftrightarrow-2sin2x.sinx+1-2sin^2x-1=0\)

\(\Leftrightarrow2sin^2x.cosx+sin^2x=0\)

\(\Leftrightarrow sin^2x\left(2cosx+1\right)=0\)

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

\(\Leftrightarrow...\)

c.

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

c. ĐKXĐ: ...

\(tanx=\frac{3}{tanx}\)

\(\Leftrightarrow tan^2x=3\)

\(\Rightarrow tanx=\pm\sqrt{3}\)

\(\Rightarrow x=\pm\frac{\pi}{3}+k\pi\)

d.

\(2sin^2x+1-2sin^2x=2\)

\(\Leftrightarrow1=2\) (vô lý)

Vậy pt vô nghiệm

a. ĐKXĐ: ...

\(cot\left(2\pi-\frac{\pi}{3}-3x\right)=tan\left(2x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow cot\left(-3x-\frac{\pi}{3}\right)=tan\left(2x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow tan\left(3x+\frac{5\pi}{6}\right)=tan\left(2x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow3x+\frac{5\pi}{6}=2x+\frac{\pi}{3}+k\pi\)

\(\Leftrightarrow...\)

b. ĐKXĐ: \(x\ne\frac{k\pi}{2}\)

\(\frac{cosx.cos2x}{sinx.sin2x}=-1\)

\(\Leftrightarrow cosx.cos2x=-sinx.sin2x\)

\(\Leftrightarrow cosx.cos2x+sinx.sin2x=0\)

\(\Leftrightarrow cosx=0\)

\(\Leftrightarrow x=\frac{\pi}{2}+k\pi\) (ktm)

Vậy pt vô nghiệm

3.

\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow x-\frac{\pi}{3}=k\pi\)

\(\Leftrightarrow...\)

4.

\(1+\frac{1}{2}sin6x=sin^2x+cos^2x+2sinx.cosx\)

\(\Leftrightarrow\frac{1}{2}sin6x=sin2x\)

\(\Leftrightarrow sin6x-2sin2x=0\)

\(\Leftrightarrow3sin2x-4sin^32x-2sin2x=0\)

\(\Leftrightarrow sin2x-4sin^32x=0\)

\(\Leftrightarrow sin2x\left(1-4sin^22x\right)=0\)

\(\Leftrightarrow sin2x\left(2cos2x-1\right)=0\)

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

1.

\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}+\sqrt{3}cos\frac{x}{2}=0\)

\(\Leftrightarrow cos\frac{x}{2}\left(2sin\frac{x}{2}+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

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

2.

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

\(\Leftrightarrow cosx=cos\left(x-\frac{\pi}{3}\right)\)

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

\(\Rightarrow x=\frac{\pi}{6}+k\pi\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(2cos^2x-1\right)-4cosx-1=0\\sinx\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4cos^2x-4cosx-3=0\\sinx\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}cosx=\frac{3}{2}\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\\sinx\ge0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2\pi}{3}+k2\pi\)

1.

\(\Leftrightarrow cosx=\frac{\sqrt{3}}{2}\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{6}+n2\pi\end{matrix}\right.\)

Do \(0< x< 2\pi\Rightarrow\left\{{}\begin{matrix}0< \frac{\pi}{6}+k2\pi< 2\pi\\0< -\frac{\pi}{6}+n2\pi< 2\pi\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\frac{1}{12}< k< \frac{11}{12}\\\frac{1}{12}< n< \frac{13}{12}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=0\\n=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}\\x=\frac{11\pi}{6}\end{matrix}\right.\) \(\Rightarrow\sum x=\frac{\pi}{6}+\frac{11\pi}{6}=2\pi\)

2.

\(-\frac{\pi}{4}\le x\le\frac{\pi}{3}\Rightarrow-\frac{\sqrt{2}}{2}\le sinx\le\frac{\sqrt{3}}{2}\)

\(\Rightarrow0\le\left|sinx\right|\le\frac{\sqrt{3}}{2}\)

\(y_{max}=\frac{\sqrt{3}}{2}\) khi \(x=\frac{\pi}{3}\)

\(y_{min}=0\) khi \(x=0\)

1.

\(2cosx-\sqrt{3}=0\\ \Leftrightarrow cosx=\frac{\sqrt{3}}{2}\\ \Leftrightarrow x=\pm\frac{\pi}{6}+k2\pi\left(k\in Z\right)\)

Do: \(0\le x\le2\pi\)

\(\Rightarrow\left\{{}\begin{matrix}0\le\frac{\pi}{6}+k2\pi\le2\pi\\0\le-\frac{\pi}{6}+k2\pi\le2\pi\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}-\frac{1}{12}\le k\le\frac{11}{12}\Rightarrow k=0\\\frac{1}{12}\le k\le\frac{13}{12}\Rightarrow k=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}k=0\Rightarrow x=\frac{\pi}{6}\\k=1\Rightarrow x=\frac{13\pi}{6}\end{matrix}\right.\)

Tổng các nghiệm là: \(\frac{\pi}{6}+\frac{13\pi}{6}=\frac{7\pi}{3}\)