Giải các phương trình
a) \(\dfrac{\cos2x}{\sin2x-1}=0\)
b) \(\cos\left(\sin x\right)=1\)
c) \(2\sin^2x-1+\cos3x=0\)
d) \(tan3x.tanx=1\)
e) \(\cos3x=-\cos7x\)
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NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
a.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)
\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 7 2021
b.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
5 tháng 9 2021
1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)
⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)
⇔ 2cos2x - 5cosx + 2 = 0
⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên
2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)
⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0
⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)
⇒ sin4x + cos4x = 48.sin4x . cos4x
⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x
⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)
⇔ 1 - 2sin22x = 0
⇔ cos4x = 0
⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
5 tháng 9 2021
3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)
⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)
⇔ sin2x - sin22x - (1 + cos4x) = 0
⇔ sin2x - sin22x - 2cos22x = 0
⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0
⇔ sin22x + sin2x - 2 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)
⇔ sin2x = 1
⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
4, cos5x + cos2x + 2sin3x . sin2x = 0
⇔ cos5x + cos2x + cosx - cos5x = 0
⇔ cos2x + cosx = 0
⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
⇔ \(cos\dfrac{3x}{2}=0\)
⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)
⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)
Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)
⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}
Vậy các nghiệm thỏa mãn là các phần tử của tập hợp
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
b/
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cosx+1-cos^2x+2cos^2x-1=\frac{1}{2}\)
\(\Leftrightarrow cos^2x+\frac{1}{2}cosx=0\)
\(\Leftrightarrow cosx\left(cosx+\frac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\left(\frac{sinx}{cosx}+\frac{cosx}{sinx}\right)^2+\frac{3}{sin2x}-7=0\)
\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{sinx.cosx}\right)^2+\frac{3}{sin2x}-7=0\)
\(\Leftrightarrow\left(\frac{2}{sin2x}\right)^2+\frac{3}{sin2x}-7=0\)
Đặt \(\frac{1}{sin2x}=a\Rightarrow4a^2+3a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{4}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=arcsin\left(-\frac{4}{7}\right)+k2\pi\\2x=\pi-arcsin\left(-\frac{4}{7}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
a/
\(\Leftrightarrow2cos2x.cosx+\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).cos2x=0\)
\(\Leftrightarrow2cos2x.cosx+cos^22x=0\)
\(\Leftrightarrow cos2x\left(2cosx+cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\left(1\right)\\2cosx+cos2x=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\left(2\right)\Leftrightarrow2cosx+2cos^2x-1=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}-1}{2}\\cosx=\frac{-\sqrt{3}-1}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
29 tháng 9 2020
a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)
b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2020
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 10 2020
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
a: ĐKXĐ: sin 2x<>1
=>2x<>pi/2+k2pi
=>x<>pi/4+kpi
\(\dfrac{cos2x}{sin2x-1}=0\)
=>cos2x=0
=>2x=pi/2+kpi
=>x=pi/4+kpi/2
Kết hợp ĐKXĐ, ta được:
x=3/4pi+k2pi hoặc x=7/4pi+k2pi
b: cos(sinx)=1
=>sin x=kpi
=>sin x=0
=>x=kpi
c: \(2\cdot sin^2x-1+cos3x=0\)
=>cos3x+cos2x=0
=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)
=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)
=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi
=>x=-pi+k2pi hoặc x=pi/5+k2pi/5
e: cos3x=-cos7x
=>cos3x=cos(pi-7x)
=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi
=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2