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a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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/
\(\Leftrightarrow2sinx+2sinx.cos2x-\left(1-sin2x\right)-2cosx=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)+2sinx\left(cos^2x-sin^2x\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)-2sinx\left(sinx-cosx\right)\left(sinx+cosx\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(2-2sin^2x-2sinx.cosx-sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cos^2x-2sinx.cosx-sinx+cosx\right]=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cosx\left(cosx-sinx\right)+cosx-sinx\right]=0\)
\(\Leftrightarrow-\left(sinx-cosx\right)^2\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\frac{tanx-1}{tanx+1}+cot2x=0\\ \Leftrightarrow cot2x-\frac{1-tanx\cdot tan\frac{\pi}{4}}{tanx+tan\frac{\pi}{4}}=0\\ \Leftrightarrow cot2x-cot\left(x+\frac{\pi}{4}\right)=0\)
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)
\(\Leftrightarrow\left(tanx+1\right)^2-2tanx=0\)
\(\Leftrightarrow tan^2x+1=0\left(vn\right)\)
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/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+\frac{5}{2}=4\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow2sin^2\left(x+\frac{\pi}{6}\right)+4sin\left(x+\frac{\pi}{6}\right)-\frac{7}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{-2+\sqrt{11}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{-2-\sqrt{11}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\\x=\frac{5\pi}{6}-arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
4.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=cos2x\)
\(\Leftrightarrow1-\frac{1}{2}sin^22x=cos2x\)
\(\Leftrightarrow1+1-sin^22x=2cos2x\)
\(\Leftrightarrow1+cos^22x=2cos2x\)
\(\Leftrightarrow\left(cos2x-1\right)^2=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow2x=k2\pi\)
\(\Rightarrow x=k\pi\)
3.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{1}{2}\)
\(\Leftrightarrow1-sin^22x=0\)
\(\Leftrightarrow cos^22x=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
Câu 1:
\(\Leftrightarrow sinx.cos\frac{\pi}{3}-cosx.sin\frac{\pi}{3}+2\left(cosx.cos\frac{\pi}{6}+sinx.sin\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow sinx+\frac{1}{\sqrt{3}}cosx=0\)
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cosx\)
\(tanx+\frac{1}{\sqrt{3}}=0\Rightarrow tanx=-\frac{1}{\sqrt{3}}\Rightarrow x=\frac{\pi}{6}+k\pi\)
Câu 2:
\(\Leftrightarrow1-cos6x=1+cos2x\)
\(\Leftrightarrow-cos6x=cos2x\)
\(\Leftrightarrow cos\left(\pi-6x\right)=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi-6x+k2\pi\\2x=6x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
Câu 3:
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}-4\pi\right)+cos2x=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)+cos2x=1\)
\(\Leftrightarrow cos2x+cos2x=1\)
\(\Leftrightarrow cos2x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Câu 4:
\(\sqrt{2}\left(cosx.cos\frac{3\pi}{4}+sinx.sin\frac{3\pi}{4}\right)=1+sinx\)
\(\Leftrightarrow-cosx+sinx=1+sinx\)
\(\Leftrightarrow cosx=-1\Rightarrow x=\pi+k\pi2\)
Câu 5:
Giống câu 3, chắc bạn ghi nhầm đề
\(\Leftrightarrow2\sqrt{2}cos2x+sin2x\left(cosx.cos\left(\frac{3\pi}{4}\right)-sinx.sin\left(\frac{3\pi}{4}\right)\right)-2\sqrt{2}\left(sinx+cosx\right)=0\)
\(\Leftrightarrow2\left(cos^2x-sin^2x\right)-sinx.cosx\left(cosx+sinx\right)-2\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx-2sinx-sinx.cosx-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\Rightarrow...\\2\left(cosx-sinx\right)-sinx.cosx-2=0\left(1\right)\end{matrix}\right.\)
Xét (1)
Đặt \(cosx-sinx=t\Rightarrow sinx.cosx=\frac{1-t^2}{2}\) (với \(\left|t\right|\le\sqrt{2}\))
\(\Rightarrow2t-\frac{1-t^2}{2}-2=0\Leftrightarrow t^2+4t-5=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-5\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow cosx-sinx=1\Leftrightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=1\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\Leftrightarrow...\)