Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
6.
\(\Leftrightarrow\frac{1}{2}cos6x+\frac{1}{2}cos4x=\frac{1}{2}cos6x+\frac{1}{2}cos2x+\frac{3}{2}+\frac{3}{2}cos2x+1\)
\(\Leftrightarrow cos4x=4cos2x+5\)
\(\Leftrightarrow2cos^22x-1=4cos2x+5\)
\(\Leftrightarrow cos^22x-2cos2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
7.
Thay lần lượt 4 đáp án ta thấy chỉ có đáp án C thỏa mãn
8.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{\pi}{6};\frac{\pi}{2}\right\}\)
9.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}-1\le t\le1\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow mt+\frac{t^2-1}{2}+1=0\)
\(\Leftrightarrow t^2+2mt+1=0\)
Pt đã cho có đúng 1 nghiệm thuộc \(\left[-1;1\right]\) khi và chỉ khi: \(\left[{}\begin{matrix}m\ge1\\m\le-1\end{matrix}\right.\)
10.
\(\frac{\sqrt{3}}{2}cos5x-\frac{1}{2}sin5x=cos3x\)
\(\Leftrightarrow cos\left(5x-\frac{\pi}{6}\right)=cos3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\frac{\pi}{6}=3x+k2\pi\\5x-\frac{\pi}{6}=-3x+k2\pi\end{matrix}\right.\)
7.
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow8cosx=\frac{\sqrt{3}cosx+sinx}{sinx.cosx}\)
\(\Leftrightarrow8cosx.sinx.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow4sin2x.cosx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x+2sinx=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow2sin3x=\sqrt{3}cosx-sinx\)
\(\Leftrightarrow sin3x=\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx\)
\(\Leftrightarrow sin\left(-3x\right)=sin\left(x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=x-\frac{\pi}{3}+k2\pi\\-3x=\frac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{2}\\x=-\frac{2\pi}{3}+k\pi\end{matrix}\right.\)
5.
\(sin\left(2x+\frac{\pi}{2}+2\pi\right)-2cos\left(x+\frac{\pi}{2}-4\pi\right)=1+2sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)-2cos\left(x+\frac{\pi}{2}\right)=1+2sinx\)
\(\Leftrightarrow cos2x+2sinx=1+2sinx\)
\(\Leftrightarrow cos2x=1\)
\(\Rightarrow x=k\pi\)
6.
\(sin^22x-cos^28x=sin\left(10x+\frac{\pi}{2}+8\pi\right)\)
\(\Leftrightarrow\frac{1-cos4x}{2}-\frac{1+cos16x}{2}=sin\left(10x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow-\left(cos4x+cos16x\right)=2cos10x\)
\(\Leftrightarrow-2cos10x.cos6x=2cos10x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\cos6x=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}10x=\frac{\pi}{2}+k\pi\\6x=\pi+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{6}+\frac{k\pi}{3}\end{matrix}\right.\)
e/
\(\Leftrightarrow\left(sin^2x+4sinx.cosx+3cos^2x\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+3cosx\right)-\left(sinx+3cosx\right)=0\)
\(\Leftrightarrow\left(sinx+3cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+3cosx=0\\sinx+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-3cosx\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-3\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-3\right)+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow2sinx+2sinx.cos2x-\left(1-sin2x\right)-2cosx=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)+2sinx\left(cos^2x-sin^2x\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow2\left(sinx-cosx\right)-2sinx\left(sinx-cosx\right)\left(sinx+cosx\right)-\left(sinx-cosx\right)^2=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(2-2sin^2x-2sinx.cosx-sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cos^2x-2sinx.cosx-sinx+cosx\right]=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left[2cosx\left(cosx-sinx\right)+cosx-sinx\right]=0\)
\(\Leftrightarrow-\left(sinx-cosx\right)^2\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(y=2+2cos\left(x-\frac{\pi}{6}\right)-7=2cos\left(x-\frac{\pi}{6}\right)-5\)
\(0\le x\le\pi\Rightarrow-\frac{\pi}{6}\le x-\frac{\pi}{6}\le\frac{5\pi}{6}\)
\(\Rightarrow-\frac{\sqrt{3}}{2}\le cos\left(x-\frac{\pi}{6}\right)\le1\)
\(\Rightarrow-\sqrt{3}-5\le y\le-3\)
\(y_{min}=-\sqrt{3}-5\) khi \(x=\pi\)
\(y_{max}=-3\) khi \(x=\frac{\pi}{6}\)
Nguyễn Lê Phước ThịnhPhạm Vũ Trí DũngMiyuki Misaki
giúp e vs ạ
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/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
Tại sao lại suy ra được -5≤y≤-1 vậy
1.
\(0\le cos^2\left(\frac{x}{2}-\frac{\pi}{9}\right)\le1\)
\(\Rightarrow-5\le y\le-1\)
\(y_{min}=-5\) khi \(cos\left(\frac{x}{2}-\frac{\pi}{9}\right)=0\)
\(y_{max}=-1\) khi \(cos^2\left(\frac{x}{2}-\frac{\pi}{9}\right)=1\)
2.
Hàm \(y=3-sin7x\) có chu kì \(T=\frac{2\pi}{7}\)
Hàm \(y=\frac{sin2x.cos2x}{25}=\frac{1}{50}sin4x\) có chu kì \(T=\frac{2\pi}{4}=\frac{\pi}{2}\)