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a) \(x=-45^0+k90^0,k\in\mathbb{Z}\)
b) \(x=-\dfrac{\pi}{6}+k\pi,k\in\mathbb{Z}\)
c) \(x=\dfrac{3\pi}{4}+k2\pi,k\in\mathbb{Z}\)
d) \(x=300^0+k540^0,k\in\mathbb{Z}\)
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/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
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\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\cdot\cos2x+\dfrac{1}{2}\cdot\sin2x+\sin\left(2x+\dfrac{\Pi}{6}\right)=\sqrt{2}\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)+\sin\left(2x+\dfrac{\Pi}{6}\right)=\sqrt{2}\)
\(\Leftrightarrow2\cdot\dfrac{\sin\left(2x+\dfrac{\Pi}{3}+2x+\dfrac{\Pi}{6}\right)}{2}\cdot\dfrac{\cos\left(2x+\dfrac{\Pi}{3}-2x-\dfrac{\Pi}{6}\right)}{2}=\sqrt{2}\)
\(\Leftrightarrow\sin\left(4x+\dfrac{\Pi}{2}\right)\cdot\cos\left(\dfrac{\Pi}{6}\right)=2\sqrt{2}\)
\(\Leftrightarrow\sin\left(4x+\dfrac{\Pi}{2}\right)=\dfrac{4\sqrt{6}}{3}\)
hay \(x\in\varnothing\)
a: \(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)+\sqrt{3}=0\)
=>\(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)=-\sqrt{3}\)
=>\(sin\left(x+\dfrac{\Omega}{5}\right)=-\dfrac{\sqrt{3}}{2}\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{5}=-\dfrac{\Omega}{3}+k2\Omega\\x+\dfrac{\Omega}{5}=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{8}{15}\Omega+k2\Omega\\x=\dfrac{4}{3}\Omega-\dfrac{\Omega}{5}+k2\Omega=\dfrac{17}{15}\Omega+k2\Omega\end{matrix}\right.\)
b: \(sin\left(2x-50^0\right)=\dfrac{\sqrt{3}}{2}\)
=>\(\left[{}\begin{matrix}2x-50^0=60^0+k\cdot360^0\\2x-50^0=300^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=110^0+k\cdot360^0\\2x=350^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=55^0+k\cdot180^0\\x=175^0+k\cdot180^0\end{matrix}\right.\)
c: \(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)-1=0\)
=>\(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)=1\)
=>\(tan\left(2x-\dfrac{\Omega}{3}\right)=\dfrac{1}{\sqrt{3}}\)
=>\(2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\)
=>\(2x=\dfrac{1}{2}\Omega+k2\Omega\)
=>\(x=\dfrac{1}{4}\Omega+k\Omega\)
Pt \(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{4}=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\),\(k\in Z\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{24}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
Hôm qua họ bảo toi ra lấy CCCD nma toi chưa đi, nay toi đi họ lại đang họp, liệu mai toi đi có bị ăn chửi ko, mn cho ý kiến đi :<
\(2sin\left(2x-\dfrac{\pi}{4}\right)+\sqrt{3}=0\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=sin\left(-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{4}=\pi+\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{12}+k2\pi\\2x=\dfrac{19\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{24}+k\pi\end{matrix}\right.\)