![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\text{) }sin^3x+cos^3x=sinx+cosx\\ \Leftrightarrow\left(sinx+cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)=sinx+cosx\\ \Leftrightarrow-\frac{1}{2}sin2x\left(sinx+cosx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=-cosx=sin\left(x-\frac{\pi}{2}\right)\\sin2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{2}-x+a2\pi\\2x=b\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{4}+a\pi\\x=\frac{b\pi}{2}\end{matrix}\right.\)
\(\text{b) }sin^3x+2sin^2x\cdot cosx-3cos^3x=0\\ \Leftrightarrow\left(sin^3x-cos^3x\right)+2cosx\cdot\left(sin^2x-cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(sinx\cdot cosx+1\right)+\left(sinx-cosx\right)\left(2sinx\cdot cosx+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(3sinx\cdot cosx+1+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(\frac{3}{2}sin2x+2+cos2x\right)=0\)
Với \(sinx-cosx=0\)
\(\Leftrightarrow sinx=cosx=sin\left(\frac{\pi}{2}-x\right)\\ \Leftrightarrow x=\frac{\pi}{2}-x+a2\pi\\ \Leftrightarrow x=\frac{\pi}{4}+a\pi\)
Với \(\frac{3}{2}sin2x+2+cos2x=0\)
\(\Leftrightarrow sin^22x+\left(\frac{3}{2}sin2x+2\right)^2=1\left(VN\right)\)
\(\text{c) }3cos^4x-4cos^2x\cdot sin^2x-sin^4x=0\)
Nhận thấy sinx=0 không là nghiệm pt.
Chia cả 2 vế cho sin4x ta được
\(pt\Leftrightarrow\frac{3cos^4x}{sin^4x}-\frac{4cos^2x}{sin^2x}-1=0\\ \Leftrightarrow3cot^4x-4cot^2x-1=0\\ \Leftrightarrow cot^2x=\frac{2+\sqrt{7}}{3}\\ \Leftrightarrow cotx=\pm\sqrt{\frac{2+\sqrt{7}}{3}}\\ \Leftrightarrow x=arccot\left(\pm\sqrt{\frac{2+\sqrt{7}}{3}}\right)+k2\pi\)
d) kiểm tra đề.
![](https://rs.olm.vn/images/avt/0.png?1311)
Với \(cosx=0\) không phải nghiệm
Với \(cosx\ne0\) , chia 2 vế cho \(cos^3x\) ta được:
\(tan^3x-5tan^2x-3tanx+3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-6tanx+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3-\sqrt{6}\\tanx=3+\sqrt{6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(3-\sqrt{6}\right)+k\pi\\x=arctan\left(3+\sqrt{6}\right)+k\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ĐKXĐ: ...
\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)
b.
\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)
\(\Leftrightarrow4cos^32x-2cos2x-1=0\)
Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề
c. ĐKXĐ: ...
\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)
\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Rightarrow x=...\\\frac{cosx+sinx}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cosx+sinx=sinx.cosx\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(4\sin (3x+\frac{\pi}{3})-2=0\Leftrightarrow \sin (3x+\frac{\pi}{3})=\frac{1}{2}=\sin (\frac{\pi}{6})\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{\pi}{3}=\frac{\pi}{6}+2k\pi \\ 3x+\frac{\pi}{3}=\pi-\frac{\pi}{6}+2k\pi\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=\frac{-\pi}{18}+\frac{2\pi}{3}\\ x=\frac{\pi}{6}+\frac{2\pi}{3}\end{matrix}\right.\) (k nguyên)
c)
\(\sin (x+\frac{x}{4})-1=0\Leftrightarrow \sin (\frac{5}{4}x)=1=\sin (\frac{\pi}{2})\)
\(\Rightarrow \frac{5}{4}x=\frac{\pi}{2}+2k\pi\Rightarrow x=\frac{2}{5}\pi+\frac{8}{5}k\pi \) (k nguyên)
d)
\(2\sin (2x+70^0)+1=0\Leftrightarrow \sin (2x+\frac{7}{18}\pi)=-\frac{1}{2}=\sin (\frac{-\pi}{6})\)
\(\Rightarrow \left[\begin{matrix} 2x+\frac{7}{18}\pi=\frac{-\pi}{6}+2k\pi\\ 2x+\frac{7}{18}\pi=\frac{7}{6}\pi+2k\pi\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=\frac{-5\pi}{18}+k\pi\\ x=\frac{7}{18}\pi+k\pi\end{matrix}\right.\)
f)
\(\cos 2x-\cos 4x=0\)
\(\Leftrightarrow \cos 2x=\cos 4x\Rightarrow \left[\begin{matrix} 4x=2x+2k\pi\\ 4x=-2x+2k\pi\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=k\pi\\ x=\frac{k}{3}\pi \end{matrix}\right.\) ( k nguyên)
b,e,g bạn xem lại đề, đơn vị không thống nhất.
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+5cosx+3=0\)
\(\Leftrightarrow2cos^2x-1+5cosx+3=0\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
\(3\left(1-sin^2x\right)+\left(1-sin^2x\right)sinx=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx\right)+\left(1+sinx\right)\left(sinx-sin^2x\right)=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx+sinx-sin^2x-8\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(-sin^2x-2sinx-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\-sin^2x-2sinx-5=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(sinx + sin5x) + (sin2x + sin4x) + 4sin3x = 0
⇔ 2sin3x . cos2x + 2sin3x . cosx + 4sin3x = 0
⇔ 2sin3x (cos2x + cosx + 2sin3x) = 0
⇔ \(\left[{}\begin{matrix}sin3x=0\left(1\right)\\cos2x+cosx+2sin3x=0\left(2\right)\end{matrix}\right.\)
(1) ⇔ ...
(2) ⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}+4sin\dfrac{3x}{2}.cos\dfrac{3x}{2}=0\)
⇔ \(\left[{}\begin{matrix}cos\dfrac{3x}{2}=0\left(\alpha\right)\\cos\dfrac{x}{2}+2sin\dfrac{3x}{2}=0\left(\beta\right)\end{matrix}\right.\)
Giải \(\left(\alpha\right)\) quá đơn giản
Giải \(\left(\beta\right)\)
\(2\left(3sin\dfrac{x}{2}-4sin^3\dfrac{x}{x}\right)+cos\dfrac{x}{2}=0\)
⇔ \(-8sin^3\dfrac{x}{2}+6sin\dfrac{x}{2}\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)+cos\dfrac{x}{2}.\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)=0\)
⇔ \(-2sin^3\dfrac{x}{2}+6sin\dfrac{x}{2}.cos^2\dfrac{x}{2}+sin^2\dfrac{x}{2}.cos\dfrac{x}{2}+cos^3\dfrac{x}{2}=0\)
Xét \(x=k2\pi,k\in Z\) tức \(sin\dfrac{x}{2}=0\) có thỏa mãn phương trình không, nếu có kết luận về nghiệm
Dù trường hợp trên có thỏa mãn hay không thì tiếp tục xét trường hợp nữa là \(x\ne k2\pi,k\in Z\) tức \(sin\dfrac{x}{2}\ne0\). Rồi chia cả 2 vế phương trình lằng nhằng kia cho \(sin\dfrac{x}{2}\) và đưa về phương trình bậc 3 theo cot\(\dfrac{x}{2}\)