a/ 1-sin9x+\(\sqrt{3}\)cos9x=0
b/ \(\sqrt{3}\)sin2x-2sin2x=\(\sqrt{2}\)-1
c/ sin5x+\(\sqrt{3}\)cos5x=2sin7x
d/ cos2x+sinx=\(\sqrt{3}\)(cosx-sin2x)
e/ sin2x+4sin2x+3cos2x+2=0
f/ 2sin2x+cos2x=7sinx-2cosx+4
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Những câu hỏi liên quan
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
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
4.
ĐKXĐ: \(2cos^2x+sinx-1\ne0\)
\(\Leftrightarrow-2sin^2x+sinx+1\ne0\Rightarrow\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)
Khi đó pt tương đương:
\(\Leftrightarrow\frac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{6}=-x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\left(loại\right)\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
3.
\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)
\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2020
3.
\(4sinx.cosx-2sinx+1-2cosx=0\)
\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
4.
\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)
Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 9 2020
5.
\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=x+k2\pi\\2x+\frac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
6.
\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)
\(\Leftrightarrow14sin^2x-5sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)
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 \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
5 tháng 9 2021
a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)
b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx
⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x
⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x
⇔ 4sin2x + (sinx + cosx) . sin2x = 0
⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)
⇔ sin2x = 0
c, 2cos3x = sin3x
⇔ 2cos3x = 3sinx - 4sin3x
⇔ 4sin3x + 2cos3x - 3sinx(sin2x + cos2x) = 0
⇔ sin3x + 2cos3x - 3sinx.cos2x = 0
Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình
Xét cosx ≠ 0 chia cả 2 vế cho cos3x ta được :
tan3x + 2 - 3tanx = 0
⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)
d, cos2x - \(\sqrt{3}sin2x\) = 1 + sin2x
⇔ cos2x - sin2x - \(\sqrt{3}sin2x\) = 1
⇔ cos2x - \(\sqrt{3}sin2x\) = 1
⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)
⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)
e, cos3x + sin3x = 2cos5x + 2sin5x
⇔ cos3x (1 - 2cos2x) + sin3x (1 - 2sin2x) = 0
⇔ cos3x . (- cos2x) + sin3x . cos2x = 0
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)
YP
20 tháng 8 2018
a.\(\dfrac{sin2x+cosx-\sqrt{3}\left(cos2x+sinx\right)}{2sin2x-\sqrt{3}}=1\left(1\right)\)
ĐKXĐ: sin2x≠\(\dfrac{\sqrt{3}}{2}\)
(1) ⇔ sin2x + cosx - \(\sqrt{3}\) ( cos2x + sinx) = 2sin2x - \(\sqrt{3}\)
⇔cosx - \(\sqrt{3}\) sinx = \(\sqrt{3}\) cos2x + sin2x +\(\sqrt{3}\)
⇔\(\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=sin\left(2x+\dfrac{\Pi}{3}\right)-sin\dfrac{\Pi}{3}\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2cos\left(x+\dfrac{\Pi}{3}\right)sinx\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)=2sin\left(\dfrac{\Pi}{6}-x\right)sinx\)
⇔\(sin\left(\dfrac{\Pi}{6}-x\right)\left(2sinx-1\right)=0\)
Đến đây tự giải tiếp nha nhớ đối chiếu đk.
YP
20 tháng 8 2018
b.\(\left(2cosx-1\right)cotx=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\left(1\right)\)
ĐKXĐ: sinx≠0 và cosx≠1
(1)⇔\(\left(2cosx-1\right)\dfrac{cosx}{sinx}=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\)
⇔cosx(2cosx-1)(cosx-1) = 3(cosx-1) + 2sin2x
⇔2cos3x - cos2x - 2cosx +1 = 0
⇔ (cosx-1)(cosx+1)(2cosx-1)=0
HP
5 tháng 9 2021
1.
\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
HP
5 tháng 9 2021
2.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)