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d/
\(\Leftrightarrow\frac{2}{\sqrt{29}}sinx-\frac{5}{\sqrt{29}}cosx=\frac{5}{\sqrt{29}}\)
Đặt \(cosa=\frac{2}{\sqrt{29}}\) với \(0< a< \pi\)
\(\Rightarrow sinx.cosa-cosx.sina=sina\)
\(\Leftrightarrow sin\left(x-a\right)=sina\)
\(\Rightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=\pi-a+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\frac{\sqrt{3}}{\sqrt{19}}cosx+\frac{4}{\sqrt{19}}sinx=\frac{\sqrt{3}}{\sqrt{19}}\)
Đặt \(cosa=\frac{\sqrt{3}}{\sqrt{19}}\) với \(0< a< \pi\)
\(\Rightarrow cosx.cosa+sinx.sina=cosa\)
\(\Leftrightarrow cos\left(x-a\right)=cosa\)
\(\Rightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=-a+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=k2\pi\end{matrix}\right.\)
1) a) cos7x - √3 sin7x = -√2 (a = 1; b = -√3; c = -√2)
=> a^2 + b^2 =4 > c^2 = 2
Chia 2 vế pt (*) cho \(\sqrt{a^2+b^2}=2\) ta đc:
<=> 1/2cos7x - √3/2 sin7x = -√2/2
<=> sin(π/6)cos7x - cos(π/6)sin7x = sin(-π/4)
<=> sin(π/6 - 7x) = sin(-π/4)
<=> π/6 - 7x = -π/4 + k2π
hoặc (k∈Z)
π/6 - 7x = π + π/4 + k2π
<=> x = 5π/84 + k2π/7
hoặc (k∈Z)
x = -13π/84 + k2π/7
1) b) Ta có:
* 2π/5 < x < 6π/7
<=> 2π/5 < 5π/84 + k2π/7 < 6π/7
<=> 143π/420 < k2π/7 < 67π/84
<=> 143/120 < k < 67/24
=> k ϵ {2}
=> x = 53π/84
* 2π/5 < x < 6π/7
<=> 2π/5 < -13π/84 + k2π/7 < 6π/7
<=> 233/120 < k < 85/24
=> k ϵ {2; 3}
=> x = 5π/12 ; x = 59π/84
Vậy có tất cả 3 nghiệm thỏa mãn (2π/5;6π/7) là x = 53π/84; x = 5π/12 ; x = 59π/84.
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a/ \(sinx=-\frac{\sqrt{3}}{2}=sin\left(-\frac{\pi}{3}\right)\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)
b/ \(cosx=\frac{\sqrt{3}}{2}=cos\left(\frac{\pi}{6}\right)\Rightarrow x=\pm\frac{\pi}{6}+k2\pi\)
c/ \(cosx=\frac{\sqrt{2}}{2}=cos\left(\frac{\pi}{4}\right)\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)
d/ \(tanx=-\sqrt{3}=tan\left(-\frac{\pi}{3}\right)\Rightarrow x=-\frac{\pi}{3}+k\pi\)
\(\text{a) }cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}=cos\frac{\pi}{3}\\\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+m2\pi\\x=n2\pi\end{matrix}\right.\)
\(\text{b) }pt\Leftrightarrow cos\left(3x-\frac{\pi}{3}\right)=\frac{1}{2}=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\3x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{9}+\frac{m2\pi}{3}\\x=\frac{n2\pi}{3}\end{matrix}\right.\)
\(\text{c) }pt\Leftrightarrow cos\left(4x+\frac{\pi}{5}\right)=-\frac{\sqrt{3}}{2}=cos\frac{5\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{5}=\frac{5\pi}{6}+m2\pi\\4x+\frac{\pi}{5}=-\frac{5\pi}{6}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{19\pi}{120}+\frac{m\pi}{2}\\x=-\frac{31\pi}{120}+\frac{n\pi}{2}\end{matrix}\right.\)
\(\text{d) }ĐKXĐ:cosx\ne-\frac{1}{2}\Leftrightarrow x\ne\pm\frac{2\pi}{3}+k2\pi\)
\(pt\Leftrightarrow2\left(4cosx+3\right)=5\left(2cosx+1\right)\\ \Leftrightarrow cosx=\frac{1}{2}=cos\frac{\pi}{3}\\ \Leftrightarrow x=\pm\frac{\pi}{3}+m2\pi\)
O pi/3 2pi/3 -pi/3 -2pi/3
Vậy \(x=\pm\frac{\pi}{3}+m2\pi\)
a/ ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\le\frac{3}{4}\\sinx\ge0\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{3-4cos2x}=\sqrt{2sinx}\)
\(\Leftrightarrow3-4cos2x=2sinx\)
\(\Leftrightarrow3-4\left(1-2sin^2x\right)=2sinx\)
\(\Leftrightarrow8sin^2x-2sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{4}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\sqrt{4sin^2x-1}=3sinx-1\) (\(\frac{1}{3}\le sinx\le1\))
\(\Leftrightarrow4sin^2x-1=9sin^2x-6sinx+1\)
\(\Leftrightarrow5sin^2x-6sinx+2=0\)
Phương trình vô nghiệm
c/
\(\Leftrightarrow2cos4x.sin3x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\2sin3x=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin3x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\3x=\frac{\pi}{6}+k2\pi\\3x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow6sinx+3cosx+3=sinx-2cosx+3\)
\(\Leftrightarrow sinx+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=0\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx=sin4x\)
\(\Leftrightarrow sin\left(\frac{\pi}{3}-x\right)=sin4x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-x+k2\pi\\4x=\frac{2\pi}{3}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)
b/
\(\Leftrightarrow2sinx.cosx+4sinx.cos^2x-2sinx=0\)
\(\Leftrightarrow2sinx\left(cosx+2cos^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\2cos^2x+cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
4.
\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+2sin^2x+3sinx-2=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\frac{\pi}{4}-x=-\frac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)
3.
\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)
\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
a,Pt \(\Leftrightarrow cosx-sinx=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\\x=-\dfrac{\pi}{4}-arc.cos\left(\dfrac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\) ,\(k\in Z\)
b) Pt \(\Leftrightarrow\dfrac{4}{5}cosx-\dfrac{3}{5}sinx=\dfrac{3}{5}\)
Đặt \(cosa=\dfrac{4}{5}\Rightarrow sina=\dfrac{3}{5}\)
Pttt:\(cosx.cosa-sina.sinx=\dfrac{3}{5}\)
\(\Leftrightarrow cos\left(x+a\right)=\dfrac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-a+arc.cos\left(\dfrac{3}{5}\right)+2k\pi\\x=-a-arc.cos\left(\dfrac{3}{5}\right)+2k\pi\end{matrix}\right.\)(\(k\in Z\))
Vậy...
c) Pt\(\Leftrightarrow\dfrac{3}{5}cos3x+\dfrac{4}{5}.sin3x=1\)
Đặt \(cosa=\dfrac{3}{5}\Rightarrow sina=\dfrac{4}{5}\)
Pttt:\(cos3x.cosa+sin3a.sina=1\)
\(\Leftrightarrow cos\left(3x-a\right)=1\)
\(\Leftrightarrow x=\dfrac{a}{3}+\dfrac{k2\pi}{3}\)(\(k\in Z\))
Vậy...
1)\(1+2sinx=2cosx\)
\(\Leftrightarrow cosx-sinx=\dfrac{1}{2}\)
\(\Leftrightarrow\left(cosx-sinx\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow cosx^2+sinx^2-2cosxsinx=\dfrac{1}{4}\)
\(\Leftrightarrow1-2cosxsinx=\dfrac{1}{4}\)
\(\Leftrightarrow2cosxsinx=\dfrac{3}{4}\)
\(\Leftrightarrow sin2x=\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x=arcsin\dfrac{3}{8}+k\pi\\x=\pi-arcsin\dfrac{3}{8}+k\pi\end{matrix}\right.\) \(\left(K\in Z\right)\)
b) \(4cosx-3sinx=3\)
\(\Leftrightarrow\dfrac{4}{5}cosx-\dfrac{3}{5}sinx=\dfrac{3}{5}\)
Đặt \(cosa=\dfrac{3}{5},sina=\dfrac{4}{5}\)
Khi đó:
\(sinacosx-cosasinx=\dfrac{3}{5}\)
\(\Leftrightarrow sin\left(a-x\right)=\dfrac{3}{5}\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-x=arcsin\dfrac{3}{5}+k2\pi\\a-x=\pi-arcsin\dfrac{3}{5}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=a-arcsin\dfrac{3}{5}+k2\pi\\x=a-\pi-arcsin\dfrac{3}{5}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
3)\(3cos3x+4sin3x=5\)
\(\Leftrightarrow\dfrac{3}{5}cos3x+\dfrac{4}{5}sin3x=1\)
Đặt \(sina=\dfrac{3}{5},cosa=\dfrac{4}{5}\)
khi đó: \(sinacos3x+cosasin3x=1\)
\(\Leftrightarrow sin\left(a+3x\right)=\dfrac{\pi}{2}\)
\(\Leftrightarrow3x=\dfrac{\pi}{2}-a+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}-\dfrac{1}{3}a+k\dfrac{2}{3}\pi\),\(k\in Z\)
Chúc bạn học tốt^^