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\(\frac{tanx-1}{tanx+1}+cot2x=0\\ \Leftrightarrow cot2x-\frac{1-tanx\cdot tan\frac{\pi}{4}}{tanx+tan\frac{\pi}{4}}=0\\ \Leftrightarrow cot2x-cot\left(x+\frac{\pi}{4}\right)=0\)
d/
ĐKXĐ: \(\left\{{}\begin{matrix}sin2x\ne0\\tanx\ne-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}+cot2x=0\\3tanx-\sqrt{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}-\frac{tan^2x-1}{2tanx}=0\\tanx=\frac{\sqrt{3}}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(tanx-1\right)\left(\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}\right)=0\left(1\right)\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Xét (1): \(\Leftrightarrow\left[{}\begin{matrix}tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\\\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}=0\left(2\right)\end{matrix}\right.\)
Xét (2)
\(\Leftrightarrow\left(tanx+1\right)^2-2tanx=0\)
\(\Leftrightarrow tan^2x+1=0\left(vn\right)\)
e/
\(\Leftrightarrow\left(sin^2x+4sinx.cosx+3cos^2x\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+3cosx\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+3cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+3cosx=0\\sinx+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-3cosx\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-3\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-3\right)+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow2sinx+2sinx.cos2x-\left(1-sin2x\right)-2cosx=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)+2sinx\left(cos^2x-sin^2x\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)-2sinx\left(sinx-cosx\right)\left(sinx+cosx\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(2-2sin^2x-2sinx.cosx-sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cos^2x-2sinx.cosx-sinx+cosx\right]=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cosx\left(cosx-sinx\right)+cosx-sinx\right]=0\)
\(\Leftrightarrow-\left(sinx-cosx\right)^2\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
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/
\(\Leftrightarrow sin2x\left(1+\sqrt{2}sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}sinx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sinx=-\frac{\sqrt{2}}{2}=sin\left(-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow2sin2x.cos2x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x=-sinx=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=-x+k2\pi\\4x=\pi+x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{5}\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
e/
\(sin\left(\frac{3\pi}{2}-sinx\right)=1\)
\(\Leftrightarrow\frac{3\pi}{2}-sinx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sinx=\pi+k2\pi\)
Mà \(-1\le sinx\le1\Rightarrow-1\le\pi+k2\pi\le1\)
\(\Rightarrow\) Không tồn tại k nguyên thỏa mãn
Pt đã cho vô nghiệm
f/
\(cos^2x-sin^2x+sin4x=0\)
\(\Leftrightarrow cos2x+2sin2x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
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
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 \(cos^3x\)
\(\Leftrightarrow\frac{1}{cos^2x}-tanx.\frac{1}{cos^2x}-4=0\)
\(\Leftrightarrow1+tan^2x-tanx\left(1+tan^2x\right)-4=0\)
\(\Leftrightarrow-tan^3x+tan^2x-tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-2tanx+3\right)=0\)
\(\Leftrightarrow tanx=-1\Rightarrow x=-\frac{\pi}{4}+k\pi\)
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 \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
c.
\(\Leftrightarrow tanx=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow x=-\frac{\pi}{6}+k\pi\)
d.
\(\Leftrightarrow\frac{1}{2}sin2x.cos2x=0\)
\(\Leftrightarrow\frac{1}{4}sin4x=0\)
\(\Leftrightarrow sin4x=0\)
\(\Leftrightarrow x=\frac{k\pi}{4}\)
e.
\(\Leftrightarrow4sin4x.cos4x=-\sqrt{2}\)
\(\Leftrightarrow2sin8x=-\sqrt{2}\)
\(\Leftrightarrow sin8x=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{4}+k2\pi\\8x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{32}+\frac{k\pi}{4}\\x=\frac{5\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)
a.
\(\Leftrightarrow sin2x=cosx\)
\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=\frac{1}{2}\\sin\frac{x}{2}=-2< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{3}+k2\pi\\\frac{x}{2}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k4\pi\\x=-\frac{2\pi}{3}+k4\pi\end{matrix}\right.\)