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a)Đặt \(t=sinx+cosx\);\(t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(\Leftrightarrow t^2=sin^2+2sinx.cosx+cos^2x\)
\(\Leftrightarrow t^2=1+2sinx.cosx\)
\(\Leftrightarrow\dfrac{t^2-1}{2}=sinx.cosx\)
Pttt: \(3t-4.\dfrac{t^2-1}{2}=0\) \(\Leftrightarrow-2t^2+3t+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(ktm\right)\\t=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow sinx.cosx=-\dfrac{3}{8}\) \(\Leftrightarrow2sinx.cosx=-\dfrac{3}{4}\)\(\Leftrightarrow sin2x=-\dfrac{3}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}.arc.sin\left(-\dfrac{3}{4}\right)+k\pi\\x=\dfrac{\pi}{2}-\dfrac{1}{2}.arc.sin\left(-\dfrac{3}{4}\right)+k\pi\end{matrix}\right.\), \(k\in Z\)
Vậy...
b)Pt
Đặt \(t=sinx-cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(\Leftrightarrow t^2-1=-2sinx.cosx\)
Pttt:\(12t+t^2-1=2\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-6+\sqrt{39}\left(tm\right)\\t=-6-\sqrt{39}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow cosx+sinx=-6+\sqrt{39}\)
\(\Leftrightarrow\sqrt{2}.cos\left(x-\dfrac{\pi}{4}\right)=-6+\sqrt{39}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arc.cos\left(\dfrac{-6+\sqrt{39}}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{\pi}{4}-arc.cos\left(\dfrac{-6+\sqrt{39}}{2}\right)+k2\pi\end{matrix}\right.\)\(,k\in Z\)
Vậy...(Nghiệm xấu)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)
a, \(cos^2x-cosx=0\)
\(\Leftrightarrow cosx\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)
Vậy...
b, \(2sin2x+\sqrt{2}sin4x=0\)
\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)
\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)
Vậy...
c, \(8cos^2x+2sinx-7=0\)
\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)
Vậy...
d, \(4cos^4x+cos^2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)
\(\Leftrightarrow cos2x=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)
Vậy...
e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))
\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)
\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)
Vậy...
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\(sin2x=\sqrt{3}cos2x\)
Nhận thấy cos2x=0 ko phải nghiệm, pt tương đương:
\(\frac{sin2x}{cos2x}=\sqrt{3}\Leftrightarrow tan2x=\sqrt{3}\)
\(\Rightarrow2x=\frac{\pi}{3}+k\pi\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)
b/
\(cos\left(90^0-x\right)=-sin2x=cos\left(2x+90^0\right)\)
\(\Rightarrow\left[{}\begin{matrix}90^0-x=2x+90^0+k360^0\\90^0-x=-2x-90^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k120^0\\x=-180^0+k360^0\end{matrix}\right.\)
c/ Giống câu a
\(\Leftrightarrow tanx=-\sqrt{3}\Rightarrow x=-\frac{\pi}{3}+k\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
ĐKXĐ: \(x\ne k\pi\)
\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.
3.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)
\(\Leftrightarrow\dfrac{3\left(1+cosx\right)}{1-cosx}+2\left(1+cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{3}{1-cosx}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
24.
\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)
\(y_{max}=4\)
26.
\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)
Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)
\(y_{max}=\sqrt{2}\)
b.
\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lý thuyết đồ thị:
Phương trình \(f\left(x\right)=m\) có nghiệm khi và chỉ khi \(f\left(x\right)_{min}\le m\le f\left(x\right)_{max}\)
Hoặc sử dụng điều kiện có nghiệm của pt lương giác bậc nhất (tùy bạn)
a.
\(\dfrac{\sqrt{3}}{2}\left(1-cos2x\right)+\dfrac{1}{2}sin2x=m\)
\(\Leftrightarrow\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x+\dfrac{\sqrt{3}}{2}=m\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)+\dfrac{\sqrt{3}}{2}=m\)
\(\Rightarrow\) Pt có nghiệm khi và chỉ khi:
\(-1+\dfrac{\sqrt{3}}{2}\le m\le1+\dfrac{\sqrt{3}}{2}\)
b.
\(\Leftrightarrow\dfrac{3}{2}\left(1-cos2x\right)-sin2x+m=0\)
\(\Leftrightarrow sin2x+\dfrac{3}{2}cos2x-\dfrac{3}{2}=m\)
\(\Leftrightarrow\dfrac{\sqrt{13}}{2}\left(\dfrac{2}{\sqrt{13}}sin2x+\dfrac{3}{\sqrt{13}}cos2x\right)-\dfrac{3}{2}=m\)
Đặt \(\dfrac{2}{\sqrt{13}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)
\(\Rightarrow\dfrac{\sqrt{13}}{2}sin\left(2x+a\right)-\dfrac{3}{2}=m\)
Phương trình có nghiệm khi và chỉ khi:
\(\dfrac{-\sqrt{13}-3}{2}\le m\le\dfrac{\sqrt{13}-3}{2}\)