giải phương trình
\(\sin x\sqrt{1+2\sin x}=\cos2x\)
\(\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}\)
\(3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)\)
\(\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x\)
\(\sin2x\sin4x+2\left(3\sin x-4\sin^2x+1\right)=0\)
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
1) \(4cos^24x+2\left(\sqrt{3}+\sqrt{2}\right)cos4x+\sqrt{6}=0\)
2) \(cos4x+2+sin\left(2x+\frac{3\pi}{2}\right)=2cos^2x\)
3) \(sin\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin\left(\frac{\pi}{6}-x\right)=1\)
4) \(2cos\left(4x-\frac{\pi}{3}\right)+4cos2x=-1\)
5) \(cos^22x+cos^23x=sin^2x\)
6) \(sinx+\left(\sqrt{2}-1\right)cosx=1\)
7) \(cos2x-\left(\sqrt{3}+1\right)cosx+\frac{2+\sqrt{3}}{2}=0\)
1.
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)
2.
\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)
\(\Leftrightarrow2cos^22x-cos2x=cos2x\)
\(\Leftrightarrow cos^22x-cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
3.
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow...\)
4.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
5.
\(cos^22x+\frac{1}{2}+\frac{1}{2}cos6x=\frac{1}{2}-\frac{1}{2}cos2x\)
\(\Leftrightarrow cos^22x+\frac{1}{2}\left(cos6x+cos2x\right)=0\)
\(\Leftrightarrow cos^22x+cos4x.cos2x=0\)
\(\Leftrightarrow cos2x\left(cos2x+cos4x\right)=0\)
\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
1)\(cos2x+5=2\sqrt{2}\left(2-cosx\right)sin\left(x-\frac{\pi}{4}\right)\)
2)
\(sin^2x-2sinx+2=sin^23x\)
3)
\(sinx-2sin2x-sin3x=2\sqrt{2}\)
4)
\(\left(cos4x-cos2x\right)^2=5+sin3x\)
5)
\(\sqrt{5+sin^23x=sinx+2cosx}\)
6)
\(5\left(sinx+\frac{cos3x+sin3x}{1+2sin2x}\right)=cos2x+3\)
7)
\(\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right)tan\left(\frac{\pi}{4}+x\right)}=cos^44x\)
7.
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(\frac{\pi}{4}-x\right).sin\left(\frac{\pi}{4}+x\right)\ne0\\cos\left(\frac{\pi}{4}-x\right)cos\left(\frac{\pi}{4}+x\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow cos2x\ne0\)
Phương trình tương đương:
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{2}-\frac{\pi}{4}-x\right)}=cos^44x\)
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{4}-x\right)}=cos^24x\)
\(\Leftrightarrow sin^42x+cos^42x=cos^44x\)
\(\Leftrightarrow\left(sin^22x+cos^22x\right)^2-2sin^22x.cos^22x=cos^44x\)
\(\Leftrightarrow1-\frac{1}{2}sin^24x=cos^44x\)
\(\Leftrightarrow2-\left(1-cos^24x\right)=2cos^44x\)
\(\Leftrightarrow2cos^44x-cos^24x-1=0\)
\(\Leftrightarrow\left(cos^24x-1\right)\left(2cos^24x+1\right)=0\)
\(\Leftrightarrow cos^24x-1=0\)
\(\Leftrightarrow sin^24x=0\Leftrightarrow sin4x=0\)
\(\Leftrightarrow2sin2x.cos2x=0\Leftrightarrow sin2x=0\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)
1.
\(cos2x+5=2\left(2-cosx\right)\left(sinx-cosx\right)\)
\(\Leftrightarrow2cos^2x+4=4sinx-4cosx-2sinx.cosx+2cos^2x\)
\(\Leftrightarrow2sinx.cosx-4\left(sinx-cosx\right)+4=0\)
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-4t+4=0\)
\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\left(sinx-1\right)^2+1=sin^23x\)
Ta có \(VT\ge1\) trong khi \(VP\le1\) với mọi x
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sinx-1=0\\sin^23x=1\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\)
3.
\(\Leftrightarrow-2cos2x.sinx-2sin2x=2\sqrt{2}\)
\(\Leftrightarrow cos2x.sinx+sin2x=-\sqrt{2}\)
Ta có:
\(VT^2=\left(cos2x.sinx+sin2x.1\right)^2\le\left(cos^22x+sin^22x\right)\left(sin^2x+1\right)\le1\left(1+1\right)=2\)
\(\Rightarrow VT\ge-\sqrt{2}\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sinx=1\\cos2x=sinx.sin2x\end{matrix}\right.\) (ko tồn tại x thỏa mãn)
Vậy pt vô nghiệm
1) \(sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right).tan^2x-cos^2\frac{x}{2}=0\)
2) \(tanx=sin^2x\left(c-\frac{\pi}{2010}\right)+cos^2\left(2x+\frac{\pi}{2010}\right)+sinx.sin\left(3x+\frac{\pi}{1005}\right)\)
3) \(1+2cosx\left(sinx-1\right)+\sqrt{2}sinx+4cosx.sin^2\frac{x}{2}=0\)
4) \(3cos4x-8cos^6x+2cos4x=3\)
5) \(1+sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)\)
6) \(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-4\sqrt{3}cos^2x.sinx.cos2x\)
7) \(\frac{tan^2x+tanx}{tan^2x+1}=\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{4}\right)\)
8) \(cos^4x+sin^4x+cos\left(x-\frac{\pi}{4}\right).sin\left(3x-\frac{\pi}{4}\right)-\frac{3}{2}=0\)
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
6.
\(sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-2\sqrt{3}cosx.sin2x.cos2x\)
\(\Leftrightarrow sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-\sqrt{3}cosx.sin4x\)
\(\Leftrightarrow sin4x\left(sinx+\sqrt{3}cosx\right)=\sqrt{2}sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin4x\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin4x.sin\left(x+\frac{\pi}{3}\right)-\frac{\sqrt{2}}{2}sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left(sin4x-\frac{\sqrt{2}}{2}\right)sin\left(x+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải các phương trình sau:
a, sinx+cosx+1+sin2x+cos2x=0
b, sinx(1+cos2x)+sin2x=1+cos2x
c, \(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
d, sin4x+cos4x=\(\frac{7}{8}cot\left(x+\frac{\pi}{3}\right)cot\left(\frac{\pi}{6}-x\right)\)
@Nguyễn Việt Lâm giúp em với ạ
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b.
\(sinx\left(1+2cos^2x-1\right)+2sinx.cosx=1+2cos^2x-1\)
\(\Leftrightarrow cos^2x.sinx+sinx.cosx-cos^2x=0\)
\(\Leftrightarrow cosx\left(sinx.cosx+sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\sinx.cosx+sinx-cosx=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(sinx-cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{1-t^2}{2}+t=0\)
\(\Leftrightarrow-t^2+2t+1=0\Rightarrow\left[{}\begin{matrix}t=1-\sqrt{2}\\t=1+\sqrt{2}>\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1-\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\\x=\frac{5\pi}{4}-arcsin\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
Giair các pt lượng giác sau:
1) \(sin\left(x-\frac{\pi}{4}\right)\left(2cos+\sqrt{2}\right)tan2x=0\)
2) \(tan2x.sinx+3\left(sin-\sqrt{3}tan2x\right)-3\sqrt{3}=0\)
3) \(\frac{cos2x}{sin\left(x+\frac{3\pi}{4}\right)}=\frac{sin\left(x+\frac{3\pi}{4}\right)}{cos2x}\)
4) \(\left(\frac{tanx-1}{tanx+1}+cot2x\right)\left(3tan-\sqrt{3}\right)=0;0< x< \pi\)
a/ ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\2cosx+\sqrt{2}=0\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\cosx=cos\left(\frac{3\pi}{4}\right)\\2x=k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=\frac{3\pi}{4}+k2\pi\left(l\right)\\x=-\frac{3\pi}{4}+k2\pi\left(l\right)\\x=\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x=\frac{k\pi}{2}\)
b/
ĐKXĐ: \(x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\Leftrightarrow tan2x.sinx+3sinx-\sqrt{3}tan2x-3\sqrt{3}=0\)
\(\Leftrightarrow sinx\left(tan2x+3\right)-\sqrt{3}\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\right)\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\sqrt{3}>1\left(vn\right)\\tan2x=-3\end{matrix}\right.\)
\(\Rightarrow2x=arctan\left(-3\right)+k\pi\)
\(\Rightarrow x=\frac{arctan\left(-2\right)}{2}+\frac{k\pi}{2}\)
c/
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x+\frac{3\pi}{4}\right)\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\frac{3\pi}{4}\ne k\pi\\2x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne-\frac{3\pi}{4}+k\pi\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(cos^22x=sin^2\left(x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{3\pi}{2}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{3\pi}{2}\right)=cos\left(2x+\frac{\pi}{2}\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=2x+\frac{\pi}{2}+k2\pi\\4x=-2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)
GPT
a) \(sin\left(2x+1\right)+cos\left(3x-1\right)=0\)
b) \(sin\left(2x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{4}\right)\)
c) \(sin\left(3x+\frac{2\pi}{3}\right)+sin\left(x-\frac{7\pi}{5}\right)=0\)
d) \(cos\left(4x+\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{4}\right)=0\)
a.
\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)
\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)
d.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
Giải các phương trình sau:
a) \(\cos \left( {3x - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}\);
b) \(2{\sin ^2}x - 1 + \cos 3x = 0\);
c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\).
a) \(\cos \left( {3x - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} = - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi + k2\pi }\\{3x = - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)
\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)
b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} = - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi + k2\pi }\\{x = - \pi + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)
1) \(\frac{1}{\cos x}+\frac{1}{\sin2x}=\frac{2}{\sin4x}\)
2) \(\cos3x\cdot\tan5x=\sin7x\)
3) \(\tan5x\cdot\tan2x=1\)
4) \(4\cos x-2\cos2x-\cos4x=1\)
5) \(\sin\left(2x+\frac{5\pi}{2}\right)-2\cos\left(x-\frac{7\pi}{2}\right)=1+2\sin x\)
6) \(\sin^22x-\cos^28x=\sin\left(\frac{17\pi}{2}+10x\right)\)
7) \(8\cos x=\frac{\sqrt{3}}{\sin x}+\frac{1}{\cos x}\)
1.
DKXĐ: \(sin4x\ne0\)
\(\Leftrightarrow\frac{4sinx.cos2x}{sin4x}+\frac{2cos2x}{sin4x}=\frac{2}{sin4x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx\left(1-2sin^2x\right)+1-2sin^2x=1\)
\(\Leftrightarrow sinx\left(1-2sin^2x-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\-2sin^2x-sinx+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
2.
ĐKXĐ: ...
\(\Leftrightarrow\frac{cos3x.sin5x}{cos5x}=sin7x\)
\(\Leftrightarrow cos3x.sin5x=sin7x.cos5x\)
\(\Leftrightarrow sin8x+sin2x=sin12x+sin2x\)
\(\Leftrightarrow sin8x=sin12x\)
\(\Leftrightarrow\left[{}\begin{matrix}12x=8x+k2\pi\\12x=\pi-8x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
Ở nghiệm đầu tiên loại các giá trị k lẻ do đó nghiệm của pt là:
\(\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
3.
ĐKXĐ: ...
\(\Leftrightarrow tan5x=\frac{1}{tan2x}\)
\(\Leftrightarrow tan5x=cot2x\)
\(\Leftrightarrow tan5x=tan\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow5x=\frac{\pi}{2}-2x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{14}+\frac{k\pi}{7}\)
