Giải PT:
sin2x + 2tanx = 0
bài 1: giải pt
a,\(\frac{cos\left(cos+2sinx\right)+3sinx\left(sinx+\sqrt{2}\right)}{sin2x-1}=1\)
b,\(\frac{sin^22x-2}{sin^22x-4cos^2x}=tan^2x\)
c, \(\frac{1+sin2x+cos2x}{1+cot^2x}=\sqrt{2}sinxsin2x\)
d, \(2tanx+cotx=2sin2x+\frac{1}{sin2x}\)
giải pt: sin2x - cos3x = 0
\(sin2x-cos3x=0\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{2}\right)-cos3x=0\)
\(\Leftrightarrow-2sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right).sin\left(-\dfrac{x}{2}-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(\dfrac{5x}{2}-\dfrac{\pi}{4}\right)=0\\sin\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5x}{2}-\dfrac{\pi}{4}=k\pi\\\dfrac{x}{2}+\dfrac{\pi}{4}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
tanx/2cosx-sin2x=0 giải pt
Giải pt: 1/sinx + 1/sin2x + 1/sin4x =0
ĐKXĐ: \(\left\{{}\begin{matrix}sinx< >0\\sin2x< >0\\sin4x< >0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >k\Omega\\2x< >k\Omega\\4x< >k\Omega\end{matrix}\right.\Leftrightarrow x\ne\dfrac{k\Omega}{4}\)
\(\dfrac{1}{sinx}+\dfrac{1}{sin2x}+\dfrac{1}{sin4x}=0\)
=>\(\dfrac{1}{sinx}+cotx+\dfrac{1}{sin2x}+cot2x+\dfrac{1}{sin4x}+cot4x=cotx+cot2x+cot4x\)
=>\(\dfrac{1+cosx}{sinx}+\dfrac{1+cos2x}{sin2x}+\dfrac{1+cos4x}{sin4x}=cotx+cot2x+cot4x\)
=>\(\dfrac{2\cdot cos^2\left(\dfrac{x}{2}\right)}{2\cdot sin\left(\dfrac{x}{2}\right)\cdot cos\left(\dfrac{x}{2}\right)}+\dfrac{2\cdot cos^2x}{2\cdot sinx\cdot cosx}+\dfrac{2\cdot cos^22x}{2\cdot sin2x\cdot cos2x}=cotx+cot2x+cot4x\)
=>\(\dfrac{cos\left(\dfrac{x}{2}\right)}{sin\left(\dfrac{x}{2}\right)}+\dfrac{cosx}{sinx}+\dfrac{cos2x}{sin2x}=cotx+cot2x+cot4x\)
=>\(cot\left(\dfrac{x}{2}\right)+cotx+cot2x=cotx+cot2x+cot4x\)
=>\(cot4x=cot\left(\dfrac{x}{2}\right)\)
=>\(\left\{{}\begin{matrix}4x=\dfrac{x}{2}+k\Omega\\4x< >k\Omega\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{7}{2}x=k\Omega\\x< >\dfrac{k\Omega}{4}\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{7}k\Omega\)
Giải pt: 1/sinx + 1/sin2x + 1/sin4x =0
\(\dfrac{1}{sinx}+\dfrac{1}{sin2x}+\dfrac{1}{sin4x}=0\)
\(\dfrac{1}{sinx}+cotx+\dfrac{1}{sin2x}+cot2x+\dfrac{1}{sin4x}+cot4x=cotx+cot2x+cot4x\)
\(\dfrac{1+cosx}{sinx}+\dfrac{1+cos2x}{sin2x}+\dfrac{1+cos4x}{sin4x}=cotx+cot2x+cot4x\)
\(\dfrac{2cos^2\dfrac{x}{2}}{2sin\dfrac{x}{2}.cos\dfrac{x}{2}}+\dfrac{2cos^2x}{2sinx.cosx}+\dfrac{2cos^22x}{2sin2x.cos2x}=cotx+cot2x+cot4x\)
\(\dfrac{cos\dfrac{x}{2}}{sin\dfrac{x}{2}}+\dfrac{cosx}{sinx}+\dfrac{cos2x}{sin2x}=cotx+cot2x+cot4x\)
\(cot\dfrac{x}{2}+cotx+cot2x=cotx+cot2x+cot4x\)
\(cot\dfrac{x}{2}=cot4x\)
\(\Rightarrow\dfrac{x}{2}=4x+k\text{π}\)
\(\Leftrightarrow x=-\dfrac{k2\text{π}}{7}\)
Giải pt
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(sin2x-cos2x+3sinx-cosx-1=0\)
1.
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)
\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)
Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)
\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
2.
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)
\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
giải pt cos(x+bi/3)+sin2x=0
\(2tanx+cotx=2sin2x+\frac{1}{sin2x}\)
\(\frac{2.sinx}{cosx}+\frac{cosx}{sinx}-4sinx.cosx-\frac{1}{2sinx.cosx}=0\) (Điều kiện \(x\ne\frac{k\pi}{2}\))
\(\Leftrightarrow\frac{4sin^2x+2cos^2x-8sin^2x.cos^2x-1}{2sinx.cosx}=0\)
\(\Leftrightarrow2sin^2x-8sin^2x.cos^2x+2\left(sin^2x+cos^2x\right)-1=0\)
\(\Leftrightarrow2sin^2x-8sin^2x.\left(1-sin^2x\right)+2-1=0\)
\(\Leftrightarrow8sin^4x-6sin^2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{4}\\sin^2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1-cos2x}{2}=\frac{1}{4}\\\frac{1-cos2x}{2}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{1}{2}=cos\frac{\pi}{3}\\cos2x=0=cos\frac{\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{4}\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
giao với điều kiện \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
giải pt: \(2cos2x+sin2x=0\)
\(pt\Leftrightarrow sin2x=-2cosx\\ \text{Mà }sin^22x+cos^22x=1\\ \Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{1}{5}\\cos2x=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{arccos\left(\frac{1}{5}\right)}{2}+m\pi\\x=\pm\frac{arccos\left(-\frac{1}{5}\right)}{2}+n\pi\end{matrix}\right.\)
Có 2 cách giải bài này:
Cách 1.
Nhận thấy \(cos2x=0\) không phải nghiệm, chia 2 vế cho \(cos2x\) ta được:
\(2+\frac{sin2x}{cos2x}=0\Leftrightarrow2+tan2x=0\Rightarrow tan2x=-2\)
Đặt \(tana=-2\Rightarrow tan2x=tana\)
\(\Rightarrow2x=a+k\pi\Rightarrow x=\frac{a}{2}+\frac{k\pi}{2}\)
(Hoặc sử dụng trực tiếp \(2x=arctan\left(-2\right)+k\pi\Rightarrow x=\frac{arctan\left(-2\right)}{2}+\frac{k\pi}{2}\))
Cách 2:
Với dạng \(a.sint+b.cost=c\) thì cách giải chung là chia 2 vế cho \(\sqrt{a^2+b^2}\) , khi đó 2 hệ số \(\frac{a}{\sqrt{a^2+b^2}}\) và \(\frac{b}{\sqrt{a^2+b^2}}\) có tổng bình phương bằng 1 nên có thể đặt thành sin, cos và sử dụng công thức lượng giác
Chia 2 vế cho \(\sqrt{5}\) ta được:
\(\frac{1}{\sqrt{5}}sin2x+\frac{2}{\sqrt{5}}cos2x=0\) (để ý rằng \(\left(\frac{1}{\sqrt{5}}\right)^2+\left(\frac{2}{\sqrt{5}}\right)^2=1\) là 1 tính chất cơ bản của sin, cos)
Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{5}}=cosa\\\frac{2}{\sqrt{5}}=sina\end{matrix}\right.\) ta được
\(sin2x.sina+cos2x.cosa=0\)
\(\Leftrightarrow sin\left(2x+a\right)=0\)
\(\Rightarrow2x+a=k\pi\Rightarrow x=-\frac{a}{2}+\frac{k\pi}{2}\)
giải PT :
tanx - sin2x - cos2x+ 2(2cosx -1/cosx )= 0