giai pt
a) \(cos^3x-sin^3x=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
b) \(\frac{1-tanx}{1+tanx}=1+2sinx\)
c) \(\left(1+tanx\right)sin^2x=3sinx\left(cosx-sinx\right)+3\)
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)
\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)
\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)
\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)
\(\Leftrightarrow cosx=sin^2x-cos^2x\)
\(\Leftrightarrow cosx=1-2cos^2x\)
\(\Leftrightarrow2cos^2x+cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
b/ ĐKXĐ: \(cos2x\ne0\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
\(6sinx-2cos^3x=\frac{10sin2x.cos2x.sinx}{2cos2x}\)
\(\Leftrightarrow6sinx-2cos^3x=5sin2x.sinx\)
\(\Leftrightarrow3sinx-cos^3x=5cosx.sin^2x\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(3tanx\left(1+tan^2x\right)-1=5tan^2x\)
\(\Leftrightarrow3tan^3x-5tan^2x+3tanx-1=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(3tan^2x-2tanx+1\right)=0\)
\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\) (ko thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
d/
\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(sinx+cosx\right)-4cos^3x\left(sin^2x+cos^2x+2sinx.cosx\right)=0\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)^2-4cos^3x\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left(cosx-sinx-4cos^3x\right)\left(sinx+cosx\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx-4cos^3x=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x+\frac{\pi}{4}=k\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
Xét \(\left(2\right)\), nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)
\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)
\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)
HP
8 tháng 2 2022
a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)
\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)
\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)
\(\Leftrightarrow-cos\left(3x-\dfrac{\pi}{2}\right)+\sqrt{3}cos^3x.\dfrac{cos^2x-3sin^2x}{cos^2x}=-2sinx\)
\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)
\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)
\(\Leftrightarrow-sin3x+\sqrt{3}cos3x=-2sinx\)
\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)
\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)-sinx=0\)
\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
Đối chiếu điều kiện ta được:
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2022
(Giả sử chọn k=-1)
Đặt \(u_n=v_n-1\Rightarrow v_{n+1}-1=\dfrac{5\left(v_n-1\right)+4}{v_n-1+2}=\dfrac{5v_n-1}{v_n+1}\)
\(\Rightarrow v_{n+1}=1+\dfrac{5v_n-1}{v_n+1}=\dfrac{6v_n}{v_n+1}\)
Mục đích chỉ cần biến đổi tới đây, sau đó nghịch đảo 2 vế:
\(\Rightarrow\dfrac{1}{v_{n+1}}=\dfrac{v_n+1}{6v_n}=\dfrac{1}{6v_n}+\dfrac{1}{6}\)
Đặt \(\dfrac{1}{v_n}=x_n\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{1}{v_1}=\dfrac{1}{u_1+1}=\dfrac{1}{6}\\x_{n+1}=\dfrac{1}{6}x_n+\dfrac{1}{6}\end{matrix}\right.\)
Rồi đó, đưa về dãy cơ bản \(\Rightarrow x_{n+1}-\dfrac{1}{5}=\dfrac{1}{6}\left(x_n-\dfrac{1}{5}\right)\)
Đặt \(x_n-\dfrac{1}{5}=y_n\Rightarrow\left\{{}\begin{matrix}y_1=x_1-\dfrac{1}{5}=-\dfrac{1}{30}\\y_{n+1}=\dfrac{1}{6}y_n\end{matrix}\right.\)
\(\Rightarrow y_n=-\dfrac{1}{30}\left(\dfrac{1}{6}\right)^{n-1}\Rightarrow x_n=y_n+\dfrac{1}{5}=-\dfrac{1}{30}.\left(\dfrac{1}{6}\right)^{n-1}+\dfrac{1}{5}\)
\(\Rightarrow v_n=\dfrac{1}{x_n}=...\Rightarrow u_n=v_n-1=\dfrac{1}{x_n}-1=...\)
Cách này là cách cơ bản, có hướng làm cố định để đưa về các dãy quen thuộc
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
HN
29 tháng 4 2020
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)
=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)
d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)
\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)
=\(\frac{1}{cosx.sinx}=VP\)
e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)
c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)
=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)
Đây nha bạn
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
d/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)
\(\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
c/
ĐKXĐ: \(sin2x\ne0\)
\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)
\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)
\(\Leftrightarrow1-cosx=sin^2x\)
\(\Leftrightarrow1-cosx=1-cos^2x\)
\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)
Đặt \(x+\frac{\pi}{4}=t\Rightarrow x=t-\frac{\pi}{4}\)
Pt trở thành:
\(sin^3t=\sqrt{2}sin\left(t-\frac{\pi}{4}\right)\)
\(\Leftrightarrow sin^3t=sint-cost\)
\(\Leftrightarrow sint-sin^3t-cost=0\)
\(\Leftrightarrow sint\left(1-sin^2t\right)-cost=0\)
\(\Leftrightarrow sint.cos^2t-cost=0\)
\(\Leftrightarrow cost\left(sint.cost-1\right)=0\)
\(\Leftrightarrow cost\left(\frac{1}{2}sin2t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=2>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cos\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
c/
ĐKXĐ: ...
Chia 2 vế cho \(cos^2x\) ta được:
\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^3x+tan^2x=3tanx-3tan^2x+3+3tan^2x\)
\(\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)