khi m≠+\(\sqrt{2}\) ≠-\(\sqrt{2}\) phương trình \(\dfrac{m\sin x-2}{m-2cosx}=\dfrac{m\cos x-2}{m-2\sin x}\) có bao nhiêu nghiệm thuộc đoạn \([20\pi;30\pi]\)
Tìm m để hàm số \(y=\sqrt{\dfrac{m-\sin x-\cos x-2\sin x\cos x}{\sin^{2017}x-\cos^{2019}x+\sqrt{2}}}\) xác định với mọi \(x\in[-\dfrac{\pi}{2};\dfrac{\pi}{2}]\)
câu 1 :Tìm x thuộc (\(\dfrac{2\pi}{5},\dfrac{6\pi}{7}\))thoã mãn PT
cos 7x - \(\sqrt{3}\) sin 7x = -2
câu 2:cho PT ;2 sin 2 x -sin x cos x - cos 2x = m
1. Tìm m để PT có nghiệm
Câu 1:
\(cos7x-\sqrt{3}sin7x=-2\\ \Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=-1\\ \Leftrightarrow7x+\dfrac{\pi}{3}=-\pi+k2\pi\\ \Leftrightarrow x=-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\)
Vì \(x\in[\dfrac{2\pi}{5};\dfrac{6\pi}{7}]\)
\(\Rightarrow\dfrac{2\pi}{5}\le x\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{2\pi}{5}\le-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{31}{15}\le k\le\dfrac{11}{3}\)
Vì \(k\in Z\) nên \(k=3\)
Vậy \(x\) cần tìm là \(\dfrac{2\pi}{3}\)
Câu 2:
\(2sin^2x-sinxcosx-cos^2x=m\\ \Leftrightarrow2\dfrac{1-cos2x}{2}-\dfrac{1}{2}s\text{in2}x-\dfrac{1+cos2x}{2}=m\\ \Leftrightarrow3cos2x+s\text{in2}x=1-2m\)
Điều kiện để phương trình có nghiệm là:
\(3^2+1^2\ge\left(1-2m\right)^2\\ \Leftrightarrow4m^2-4m-9\le0\\ \Leftrightarrow\dfrac{1-\sqrt{10}}{2}\le m\le\dfrac{1+\sqrt{10}}{2}\)
Phương trình \(sin\left(x^2-5x\right)=\dfrac{-\sqrt{3}}{2}\) có bao nhiêu nghiệm thuộc \(\left[0;\dfrac{\pi}{2}\right]\)
tìm m pt : 2 ( sin x +cos x) + sin2x +m2 -4\(\sqrt{2}\) m =0 có nghiệm x thuộc (0;\(\dfrac{\pi}{2}\))
1) sin2x + 2cosx = 0
2) sin(2x -10*) = \(\dfrac{1}{2}\) (-120* <x< 90*)
3) cos(2x+10*)= \(\dfrac{\sqrt{2}}{2}\)(-180*<x<180*)
4) \(\sin^2\left(5x+\dfrac{2\pi}{5}\right)-\cos^2\)(\(\dfrac{x}{4}-\pi\)) =0
1.
\(\Leftrightarrow2sinx.cosx+2cosx=0\)
\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
2.
\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)
Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)
3. Làm tương tự câu 2
4.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.3 .giải phương trình
d) sin 8x - cos 6x = \(\sqrt{3}\)(sin 6x + cos 8x)
3.4 .giải pt
a) 2sin(\(x+\dfrac{\pi}{4}\)) + 4 sin (\(x-\dfrac{\pi}{4}\)) = \(\dfrac{3\sqrt{5}}{2}\)
b)3 sin (x-\(\dfrac{\pi}{3}\)) + 4 sin (x +\(\dfrac{\pi}{6}\)) + 5 sin(5x +\(\dfrac{\pi}{6}\)) = 0
3.9 a) 8sin x =\(\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\)
b)\(2\sqrt{sinx}=\dfrac{\sqrt{3}tanx}{2\sqrt{sinx}-1}-1\)
mọi người ơi giúp mình với mình sắp phải kiểm tra rồi
3.3 d)
\(\sin8x-\cos6x=\sqrt{3}\left(\sin6x+\cos8x\right)\\ \Leftrightarrow\sin8x-\sqrt{3}\cos8x=\sqrt{3}\sin6x+\cos6x\\ \Leftrightarrow\sin\left(8x-\dfrac{\pi}{3}\right)=\sin\left(6x+\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\pi-\left(6x+\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{7}\end{matrix}\right.\)
3.4 a)
\(2sin\left(x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(-x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \)
Chia hai vế cho \(\sqrt{2^2+4^2}=2\sqrt{5}\)
Ta được:
\(\dfrac{1}{\sqrt{5}}cos\left(x-\dfrac{\pi}{4}\right)+\dfrac{2}{\sqrt{5}}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3}{4}\\ \)
Gọi \(\alpha\) là góc có \(cos\alpha=\dfrac{1}{\sqrt{5}}\)và \(sin\alpha=\dfrac{2}{\sqrt{5}}\)
Phương trình tương đương:
\(cos\left(x-\dfrac{\pi}{4}-\alpha\right)=\dfrac{3}{4}\\ \Leftrightarrow x=\pm arscos\left(\dfrac{3}{4}\right)+\dfrac{\pi}{4}+\alpha+k2\pi\)
Giải các pt
a) \(\sqrt{2}\sin\left(2x+\dfrac{\pi}{4}\right)=3\sin x+\cos x+2\)
b) \(\dfrac{\left(2-\sqrt{3}\right)\cos x-2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2\cos x-1}=1\)
c) \(2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x\right)\sin x=1\)
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
c.
\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)
\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)
\(\Leftrightarrow...\)
Tìm m để phương trình sau có nghiệm:
\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt{3}.sin2x-cos2x\)
\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt[]{3}sin2x-cos2x\)
\(\Leftrightarrow4.\left(-\dfrac{1}{2}\right)\left[sin\left(x+\dfrac{\pi}{3}+x-\dfrac{\pi}{6}\right)+sin\left(x+\dfrac{\pi}{3}-x+\dfrac{\pi}{6}\right)\right]=m^2+2.\left[\dfrac{\sqrt[]{3}}{2}.sin2x-\dfrac{1}{2}.cos2x\right]\)
\(\Leftrightarrow2\left[sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(2x-\dfrac{\pi}{6}\right)\right]=m^2+2\)
\(\Leftrightarrow2.2sin2x.cos\dfrac{\pi}{6}=m^2+2\)
\(\Leftrightarrow2.2sin2x.\dfrac{\sqrt[]{3}}{2}=m^2+2\)
\(\Leftrightarrow2\sqrt[]{3}sin2x.=m^2+2\)
\(\Leftrightarrow sin2x.=\dfrac{m^2+2}{2\sqrt[]{3}}\)
Phương trình có nghiệm khi và chỉ khi
\(\left|\dfrac{m^2+2}{2\sqrt[]{3}}\right|\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m^2+2}{2\sqrt[]{3}}\ge-1\\\dfrac{m^2+2}{2\sqrt[]{3}}\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2\ge-2\left(1+\sqrt[]{3}\right)\left(luôn.đúng\right)\\m^2\le2\left(1-\sqrt[]{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow-\sqrt[]{2\left(1-\sqrt[]{3}\right)}\le m\le\sqrt[]{2\left(1-\sqrt[]{3}\right)}\)
Có bao nhiêu giá trị nguyên của m thuộc đoạn [-10;10] để phương trình \(sin\left(x-\dfrac{\pi}{3}\right)-\sqrt{3}cos\left(x-\dfrac{\pi}{3}\right)=2m\) vô nghiệm.
\(sin\left(x-\dfrac{\pi}{3}\right)-\sqrt{3}cos\left(x-\dfrac{\pi}{3}\right)=2m\)
\(\Leftrightarrow\dfrac{1}{2}sin\left(x-\dfrac{\pi}{3}\right)-\dfrac{\sqrt{3}}{2}cos\left(x-\dfrac{\pi}{3}\right)=m\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right).cos\dfrac{\pi}{3}-cos\left(x-\dfrac{\pi}{3}\right).sin\dfrac{\pi}{3}=m\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}-\dfrac{\pi}{3}\right)=m\)
\(\Leftrightarrow sin\left(x-\dfrac{2\pi}{3}\right)=m\)
Do \(-1\le sin\left(x-\dfrac{2\pi}{3}\right)\le1\) \(\Rightarrow\) pt vô nghiệm khi và chỉ khi \(\left|m\right|>1\)
\(\Rightarrow\left[{}\begin{matrix}-10\le m< -1\\1< m\le10\end{matrix}\right.\) \(\Rightarrow\) có 18 giá trị nguyên của m để pt đã cho vô nghiệm