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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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
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 \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
d/
\(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}cos\left(\frac{x}{5}-\frac{\pi}{12}\right)-\frac{\sqrt{3}}{2}sin\left(\frac{x}{5}-\frac{\pi}{12}\right)\right)=sin\left(\frac{x}{5}+\frac{2\pi}{3}\right)-sin\left(\frac{3x}{5}+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\sqrt{2}cos\left(\frac{x}{5}-\frac{\pi}{12}+\frac{\pi}{3}\right)=2cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)sin\left(\frac{\pi}{4}-\frac{x}{5}\right)\)
\(\Leftrightarrow cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=\sqrt{2}cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)cos\left(\frac{x}{5}-\frac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{x}{5}-\frac{\pi}{4}\right)=0\\cos\left(\frac{2x}{5}+\frac{5\pi}{12}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{5}-\frac{\pi}{4}=\frac{\pi}{2}+k\pi\\\frac{2x}{5}+\frac{5\pi}{12}=\frac{\pi}{4}+k2\pi\\\frac{2x}{5}+\frac{5\pi}{12}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{15\pi}{4}+k5\pi\\x=-\frac{5\pi}{12}+k5\pi\\x=-\frac{5\pi}{3}+k5\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x\)
\(\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.\)
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:
\(1-t^2-4t+4=0\)
\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
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.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
d/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+\frac{5}{2}=4\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow2sin^2\left(x+\frac{\pi}{6}\right)+4sin\left(x+\frac{\pi}{6}\right)-\frac{7}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{-2+\sqrt{11}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{-2-\sqrt{11}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\\x=\frac{5\pi}{6}-arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
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:
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
(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