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a.
Với \(cosx=0\) ko phải nghiệm
Với \(cosx\ne0\) chia 2 vế cho \(cos^2x\)
\(\Rightarrow-3tanx+tan^2x=2+2tan^2x\)
\(\Leftrightarrow tan^2x+3tanx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)
b.
Với \(cosx=0\) không phải nghiệm
Với \(cosx\ne0\) chia 2 vế cho \(cos^2x\)
\(\Rightarrow2tan^2x+tanx-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(-\dfrac{3}{2}\right)+k\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
![](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)
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)
<=> 2.sinx.cosx - 1 + 2.sin2x + 3.sinx - cosx - 1 = 0
<=> cosx.( 2.sinx - 1 ) + ( 2.sinx - 1 ).( sinx + 2) = 0
<=> (2.sinx - 1 ).( cosx + sinx + 2 ) = 0
<=> \(\left[{}\begin{matrix}2.sinx-1=0\\cosx+sinx=2\left(VN\right)\end{matrix}\right.\)
<=> sinx = \(\dfrac{1}{2}\)
<=> \(\left[{}\begin{matrix}x=\dfrac{\pi}{6}+2k\pi\\x=\dfrac{5\pi}{6}+2k\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
+ , cos3 x = 0 => 0 - 4 - 0 + 1 = 0 ( vô nghiệm)
+, cos3 x \(\ne\)0 , chia cả 2 vế của pt cho cos3 x , ta đc
\(\frac{\cos^3x-4sin^3x-3cosx.sin^2x+sinx}{cos^3x}=0\)
1 - \(\frac{4\sin^3x}{\cos^3x}\) - \(\frac{3\sin^2x}{cos^2x}\) + \(\frac{1}{\cos^2x}\)= 0
1 - 4 tan3x - 3 tan2x + 1 + tan2x = 0
-4 tan3x - 2tan2x + 2 = 0
=> tan x = tan \(\alpha\) ( tan \(\alpha\approx0,66\))
=> x = \(\alpha+k.\pi\)
sin6(2x)+cos6(2x)=0
<=>(sin22x)3+(cos22s)3=0
<=> (sin22x+cos2x)[(sin42x-2sin22x.cos22x+cos42x)=0
<=>sin42x-2sin22x.cos22x+cos42x=0
đến đây bạn nhóm l;ại và giải
KQ là k có nghiệm
=> PT vô nghiệm
mình cảm ơn nhé