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\(2sinx+2\sqrt{3}cosx-\sqrt{3}sin2x+cos2x=\sqrt{3}cosx+cos2x-2sinx+2\)
\(\Leftrightarrow4sinx+\sqrt{3}cosx-2\sqrt{3}sinx.cosx-2=0\)
\(\Leftrightarrow-2sinx\left(\sqrt{3}cosx-2\right)+\sqrt{3}cosx-2=0\)
\(\Leftrightarrow\left(1-2sinx\right)\left(\sqrt{3}cosx-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\cosx=\dfrac{2}{\sqrt{3}}>1\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(\sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{2}\Leftrightarrow x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{3\pi}{4}+k2\pi\left(k\in Z\right)\)
`sin(x- (pi)/4) = (pi)/2`
`<=> x - (pi)/4 = (pi)/2 + k2(pi)`
`<=> x = (3(pi))/4 + k2(pi)`.
\(\sin\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{3}+k\pi\left(k\in Z\right)\)
\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\left(k\in Z\right)\)
1.
Chắc đề là \(sin\left[\pi sin2x\right]=1?\)
\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)
Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)
\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)
Thế vào (1)
\(\Rightarrow sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)
Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)
Thế vào (2):
\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\) chắc bạn tự giải tiếp được
\(\Leftrightarrow\left[{}\begin{matrix}5x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\5x+\dfrac{\pi}{6}=\pi-\left(x-\dfrac{\pi}{3}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-\dfrac{\pi}{2}+k2\pi\\6x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\x=\dfrac{7\pi}{36}+k\dfrac{\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)