3.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ
1.
\(sin\left(sinx\right)=0\)
\(\Leftrightarrow sinx=k\pi\) (1)
Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)
\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)
Thế vào (1)
\(\Rightarrow sinx=0\Rightarrow x=n\pi\)
2.
\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)
Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)
\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\)
\(\Rightarrow\) Không tồn tại k thỏa mãn
Pt vô nghiệm
4.
\(sin2x=sin\left(2x-\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=2x-\dfrac{\pi}{2}+k2\pi\left(vn\right)\\2x=\dfrac{3\pi}{2}-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{3\pi}{8}+\dfrac{k\pi}{2}\)
5.
\(\Leftrightarrow cos\left(3\pi-2x\right)=\dfrac{\sqrt{2}}{4}\)
\(\Leftrightarrow cos2x=-\dfrac{\sqrt{2}}{4}\)
\(\Leftrightarrow2x=\pm arccos\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\)
\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(-\dfrac{\sqrt{2}}{4}\right)+k\pi\)