giải phương trình:
3tan2x + 2cos2x = \(\dfrac{3}{cos2x}\) + 2tan (x- \(\dfrac{\pi}{4}\))
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NL
0
HP
17 tháng 9 2021
d, Hàm số xác định khi:
\(\left\{{}\begin{matrix}cos\left(x+\dfrac{\pi}{4}\right)\ne0\\sinx.cosx+cos2x-3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+k\pi\\\dfrac{1}{2}sin2x+cos2x\ne3\end{matrix}\right.\)
\(\Leftrightarrow x\ne\dfrac{\pi}{4}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 7 2021
1.
\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)
\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)
\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)
\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)
2.
Đề bài thiếu, cos?x
Và x thuộc khoảng nào?
3.
\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)
\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)
\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)
4.
\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)
\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)
HP
1 tháng 6 2021
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:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
HP
1 tháng 6 2021
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 \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
6 tháng 11 2023
a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{4}+k2\Omega\\x=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
b: cos2x=cosx
=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{k2\Omega}{3}\)
c:
ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)
\(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)
=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{2}{3}\Omega+k\Omega\)
d:
ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)
=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)
=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)
\(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)
=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)
=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)
=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)
15 tháng 12 2023
1: \(P=sin^22x=1-cos^22x\)
\(=1-\left(cos2x\right)^2\)
\(=1-\left(2cos^2x-1\right)^2\)
\(=1-\left(2\cdot\dfrac{9}{16}-1\right)^2\)
\(=1-\left(\dfrac{9}{8}-1\right)^2=1-\left(\dfrac{1}{8}\right)^2=\dfrac{63}{64}\)
2:
\(cos2x-sin\left(x+\dfrac{\Omega}{3}\right)=0\)
=>\(sin\left(x+\dfrac{\Omega}{3}\right)=cos2x=sin\left(\dfrac{\Omega}{2}-2x\right)\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=\dfrac{\Omega}{2}-2x+k2\Omega\\x+\dfrac{\Omega}{3}=\Omega-\dfrac{\Omega}{2}+2x+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=\dfrac{\Omega}{6}+k2\Omega\\-x=\dfrac{1}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{18}+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{6}\Omega-k2\Omega\end{matrix}\right.\)
MA
1
AH
Akai Haruma
Giáo viên
11 tháng 9 2021
Lời giải:
$\tan (\frac{\pi}{2}+x)-3\tan ^2x=\frac{\cos 2x-1}{\cos ^2x}=\frac{2\cos ^2x-2}{\cos ^2x}=\frac{2(\cos ^2x-1)}{\cos ^2x}$
$=\frac{-2\sin ^2x}{\cos ^2x}=-2\tan ^2x$
$\Leftrightarrow \tan (x+\frac{\pi}{2})=\tan ^2x$
Dễ thấy $\tan x=0$ không thỏa mãn nên $\tan x\neq 0$. Do đó pt $\Leftrightarrow \tan ^2x=\tan [\pi +(x-\frac{\pi}{2})]=\tan (x-\frac{\pi}{2})=-\tan (\frac{\pi}{2}-x)=-\cot x =\frac{-1}{\tan x}$
$\Rightarrow \tan ^3x=-1$
$\Rightarrow \tan x=-1$
$\Rightarrow x=\frac{-\pi}{4}+k\pi$ với $k$ nguyên.
9 tháng 9 2023
\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt[]{3}sin2x-cos2x\)
\(\Leftrightarrow4.\left(-\dfrac{1}{2}\right)\left[sin\left(x+\dfrac{\pi}{3}+x-\dfrac{\pi}{6}\right)+sin\left(x+\dfrac{\pi}{3}-x+\dfrac{\pi}{6}\right)\right]=m^2+2.\left[\dfrac{\sqrt[]{3}}{2}.sin2x-\dfrac{1}{2}.cos2x\right]\)
\(\Leftrightarrow2\left[sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(2x-\dfrac{\pi}{6}\right)\right]=m^2+2\)
\(\Leftrightarrow2.2sin2x.cos\dfrac{\pi}{6}=m^2+2\)
\(\Leftrightarrow2.2sin2x.\dfrac{\sqrt[]{3}}{2}=m^2+2\)
\(\Leftrightarrow2\sqrt[]{3}sin2x.=m^2+2\)
\(\Leftrightarrow sin2x.=\dfrac{m^2+2}{2\sqrt[]{3}}\)
Phương trình có nghiệm khi và chỉ khi
\(\left|\dfrac{m^2+2}{2\sqrt[]{3}}\right|\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m^2+2}{2\sqrt[]{3}}\ge-1\\\dfrac{m^2+2}{2\sqrt[]{3}}\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2\ge-2\left(1+\sqrt[]{3}\right)\left(luôn.đúng\right)\\m^2\le2\left(1-\sqrt[]{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow-\sqrt[]{2\left(1-\sqrt[]{3}\right)}\le m\le\sqrt[]{2\left(1-\sqrt[]{3}\right)}\)
Điều kiện xác định : sin4x ≠ 0
3tan2x + 2cos2x = \(\dfrac{3}{cos2x}\) + 2 \(\dfrac{sin\left(x-\dfrac{\pi}{4}\right)}{cos\left(x-\dfrac{\pi}{4}\right)}\)
⇔ 3tan2x + 2cos2x = \(\dfrac{3}{cos2x}\) + 2 \(\dfrac{sinx-cosx}{sinx+cosx}\)
⇒ 3tan2x . cos2x + 2cos22x = 3 + 2\(\dfrac{sinx-cosx}{sinx+cosx}\).cos2x
⇒ 3sin2x + 2cos22x = 3 + 2. \(\dfrac{sinx-cosx}{sinx+cosx}\).(cosx - sinx)(cosx + sinx)
⇒ 3sin2x + 2cos22x = 3 - 2(sinx - cosx)2
⇔ 3sin2x + 2cos22x = 3 - 2 . (1 - sin2x)
⇔ 3sin2x + 2 - 2sin22x = 3 - 2 + 2sin2x
⇔ - 2sin22x + sin2x + 1 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-\dfrac{1}{2}\end{matrix}\right.\)
Loại sin2x = 1 vì khi đó cos2x = 0 (vi phạm ĐKXĐ)
⇔ sin2x = \(-\dfrac{1}{2}\)
Giải nốt nhé