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c/
\(\Leftrightarrow1-sin^22x+\sqrt{3}sin2x+sin2x=1+\sqrt{3}\)
\(\Leftrightarrow-sin^22x+\left(\sqrt{3}+1\right)sin2x-\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\sqrt{3}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
d/
\(\Leftrightarrow4\left(1-2sin^2x\right)+5sinx=4\left(3sinx-4sin^3x\right)+5\)
\(\Leftrightarrow16sin^3x-8sin^2x-7sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(4sinx+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=arcsin\left(-\frac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{4}\right)+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow3cos^2x+4sin\left(2\pi-\frac{\pi}{2}-x\right)+1=0\)
\(\Leftrightarrow3cos^2x-4sin\left(x+\frac{\pi}{2}\right)+1=0\)
\(\Leftrightarrow3cos^2x-4cosx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm arcos\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
5.
\(\Leftrightarrow sin\left(2cosx\right)=1\)
\(\Leftrightarrow2cosx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow cosx=\frac{\pi}{4}+k\pi\)
Do \(-1\le cosx\le1\Rightarrow-1\le\frac{\pi}{4}+k\pi\le1\)
Mà \(k\in Z\Rightarrow k=0\)
\(\Rightarrow cosx=\frac{\pi}{4}\)
\(\Leftrightarrow x=\pm arccos\left(\frac{\pi}{4}\right)+k2\pi\)
3.
\(\Leftrightarrow sin2x+1=2\left(\frac{1-cos2x}{2}\right)\)
\(\Leftrightarrow sin2x+cos2x=0\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow2x+\frac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)
4. ĐKXĐ; ...
\(\Leftrightarrow\frac{sinx.cos2x}{cosx.sin2x}+1=0\)
\(\Leftrightarrow sinx.cos2x+cosx.sin2x=0\)
\(\Leftrightarrow sin3x=0\)
\(\Leftrightarrow3sinx-4sin^3x=0\)
\(\Leftrightarrow3-4sin^2x=0\)
\(\Leftrightarrow3-2\left(1-cos2x\right)=0\)
\(\Leftrightarrow cos2x=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
1.
\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}\left(cos^4\frac{x}{2}-sin^4\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sinx.cosx=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow\frac{1}{2}sin2x=\frac{\sqrt{3}}{4}\)
\(\Leftrightarrow sin2x=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
3.
ĐKXĐ: ...
\(\frac{1}{cosx}+\frac{1}{2sinx.cosx}=\frac{1}{2sinx.cosx.cos2x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow cos2x-sinx=0\)
\(\Leftrightarrow1-2sin^2x-sinx=0\)
\(\Leftrightarrow2sin^2x+sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
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 \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
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
