1) Giai phuong trinh:
a) \(\sqrt{3}sin|\dfrac{\Pi}{2}-5x|-sin\left(\Pi+5x\right)+2sin2x=0\)
1) Giai phuong trinh:
a) \(\sqrt{3}sin|\dfrac{\Pi}{2}-5x|-sin\left(\Pi+5x\right)+2sin2x=0\)
Căn3/(cosx*cosx)=3*cosx + căn3
3cosx - 4sinx=5 . giúp vs ạ
\(\Leftrightarrow\dfrac{3}{5}cosx-\dfrac{4}{5}sinx=1\)
Đặt \(\dfrac{3}{5}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{4}{5}=sina\)
Phương trình trở thành:
\(cosa.cosx-sina.sinx=1\)
\(\Leftrightarrow cos\left(x-a\right)=1\)
\(\Leftrightarrow x-a=k2\pi\)
\(\Leftrightarrow x=a+k2\pi\) (\(k\in Z\))
2cos^2x -5 sinx -2 =0
\(2cos^2x-5sinx-2=0\)
\(\Rightarrow2\left(1-sin^2x\right)-5sinx-2=0\)
\(\Rightarrow-2sin^2x-5sinx=0\)
\(\Rightarrow\left\{{}\begin{matrix}sinx=0\\2sinx+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=k\pi\\sinx=-\dfrac{5}{2}\left(l\right)\end{matrix}\right.\)\(\left(k\in Z\right)\)
Giúp mình với ạ
\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sinx+\dfrac{\sqrt{2}}{2}cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
c) \(sinx+cosx=1\)
\(\Rightarrow\)\(\sqrt{2}sin\left(x+\dfrac{\pi}{2}\right)=1\) \(\Rightarrow sin\left(x+\dfrac{\pi}{2}\right)=\dfrac{1}{\sqrt{2}}\)\(=sin\dfrac{\pi}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{\pi}{2}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{2}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
\(\Leftrightarrow3\left(2cos^23x-1\right)+\left(3\sqrt{3}-2\right)cos3x+2-\sqrt{3}=0\)
\(\Leftrightarrow6cos^23x+\left(3\sqrt{3}-2\right)cos3x-1-\sqrt{3}=0\)
Phương trình bậc 2 này có nghiệm rất xấu
\(\Leftrightarrow6-6sin^2\dfrac{x}{2}+\left(8+3\sqrt{3}\right)sin\dfrac{x}{2}-6-4\sqrt{3}=0\)
\(\Leftrightarrow-6sin^2\dfrac{x}{2}+\left(8+3\sqrt{3}\right)sin\dfrac{x}{2}-4\sqrt{3}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\dfrac{x}{2}=\dfrac{\sqrt{3}}{2}\\sin\dfrac{x}{2}=\dfrac{4}{3}>1\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{\pi}{3}+k2\pi\\\dfrac{x}{2}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k4\pi\\x=\dfrac{4\pi}{3}+k4\pi\end{matrix}\right.\)
Cho cosx = -3/5 , 𝝅/2 <x<𝝅. Nêu cách tính sinx, sin (x + 𝝅/3), (x+ 𝝅/4)
\(\dfrac{\pi}{2}< x< \pi\Rightarrow sinx>0\)
\(\Rightarrow sinx=\sqrt{1-cos^2x}=\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)
\(sin\left(x+\dfrac{\pi}{3}\right)=sinx.cos\left(\dfrac{\pi}{3}\right)+cosx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{4}{5}.\dfrac{1}{2}+\left(-\dfrac{3}{5}\right).\dfrac{\sqrt{3}}{2}=\dfrac{4-3\sqrt{3}}{10}\)
\(cos\left(x+\dfrac{\pi}{4}\right)=cosx.cos\left(\dfrac{\pi}{4}\right)-sinx.sin\left(\dfrac{\pi}{4}\right)=-\dfrac{3}{5}.\dfrac{\sqrt{2}}{2}-\dfrac{4}{5}.\dfrac{\sqrt{2}}{2}=-\dfrac{7\sqrt{2}}{10}\)
\(0< a< \dfrac{\pi}{2}\Rightarrow cosa>0\)
\(\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{2\sqrt{2}}{3}\)
\(sin\left(a-\dfrac{\pi}{4}\right)=sina.cos\left(\dfrac{\pi}{4}\right)-cosa.sin\left(\dfrac{\pi}{4}\right)=\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-\dfrac{2\sqrt{2}}{3}.\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}-8}{6}\)
\(cos\left(a-\dfrac{\pi}{6}\right)=cosa.cos\left(\dfrac{\pi}{6}\right)+sina.sin\left(\dfrac{\pi}{6}\right)=\dfrac{2\sqrt{2}}{3}.\dfrac{1}{2}+\dfrac{1}{3}.\dfrac{1}{2}=\dfrac{2\sqrt{2}+1}{6}\)
1) Tai sao Sin [(x+\(\dfrac{\Pi}{2}\)) - 2π] = Sin (x+\(\dfrac{\Pi}{2}\))
Do \(sin\left(a+k2\pi\right)=sina\)