giải phương trình: \(\sin\left(3x+\dfrac{\pi}{4}\right)=\sin\left(x-\dfrac{\pi}{3}\right)\)
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HP
12 tháng 9 2021
\(\sqrt{3}cos\left(x+\dfrac{\pi}{2}\right)+sin\left(x-\dfrac{\pi}{2}\right)=2sin2x\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{2}\right)-\dfrac{1}{2}cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{2}+x\right)=sin2x\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx+sin2x=0\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)+sin2x=0\)
\(\Leftrightarrow2sin\left(\dfrac{3x}{2}+\dfrac{\pi}{12}\right).cos\left(\dfrac{\pi}{12}-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(\dfrac{3x}{2}+\dfrac{\pi}{12}\right)=0\\cos\left(\dfrac{\pi}{12}-\dfrac{x}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}+\dfrac{\pi}{12}=k\pi\\\dfrac{\pi}{12}-\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\\x=-\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
TK
14 tháng 7 2017
Đặt \(t=\dfrac{3\pi}{10}-\dfrac{x}{2}\)\(\Rightarrow\pi-3t=\dfrac{\pi}{10}+\dfrac{3\pi}{2}\)
\(pt\Leftrightarrow2sint=sin\left(\pi-3t\right)\)
\(\Leftrightarrow2sint=3sint-4sin^3t\)
\(\Leftrightarrow sint\left(1-4sin^2t\right)=0\)
\(\Leftrightarrow sint\left(2cos2t\right)=0\)
dễ nhé :3
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=x-\dfrac{\pi}{3}+k2\pi\\3x+\dfrac{\pi}{4}=\pi-\left(x-\dfrac{\pi}{3}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-7\pi}{12}+k2\pi\\4x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7\pi}{24}+k\pi\\x=\dfrac{13\pi}{48}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
\(sin\left(3x+\dfrac{\Pi}{4}\right)=sin\left(x-\dfrac{\Pi}{3}\right)\)
\(\Leftrightarrow3x+\dfrac{\Pi}{4}=x-\dfrac{\Pi}{3}+K2\Pi\)
\(\Leftrightarrow2x=-\dfrac{7\Pi}{12}+K2\Pi\)
\(\Leftrightarrow x=-\dfrac{7\Pi}{24}+K\Pi\) \(\left(K\in Z\right)\)