Giải phương trình: 1.$cos^3x.cos3x sin^3x.sin3x=\frac{\sqrt{2}}{4}$ 2.$cos^34x=cos^3x.cos3x sin^3x.sin3x$ 3.$cos^2x-4sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right) 2=0$ 4.\(sin^4x sin^4\left(x ...

TD

tran duc huy

8 tháng 8 2020

Giải phương...

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#Toán lớp 11

9

NV

Nguyễn Việt Lâm

Giáo viên

8 tháng 8 2020

6.

$\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0$

$\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0$

$\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0$

$\Leftrightarrow-3sin^22x+sin2x+4=0$

$\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.$

$\Rightarrow2x=-\frac{\pi}{2}+k2\pi$

$\Rightarrow x=-\frac{\pi}{4}+k\pi$

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NV

Nguyễn Việt Lâm

Giáo viên

8 tháng 8 2020

5.

$\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]$

$\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)$

$\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)$

$\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}$

$\Leftrightarrow sin^22x=\frac{1}{2}$

$\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.$

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QT

Quoc Tran Anh Le

Giáo viên

21 tháng 9 2023

Giải các phương trình sau:

a) $\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}$

b) $\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}$

c) $\sin 3x - \cos 5x = 0$

d) ${\cos ^2}x = \frac{1}{4}$

e) $\sin x - \sqrt 3 \cos x = 0$

f) $\sin x + \cos x = 0$

#Toán lớp 11

2

HQ

Hà Quang Minh

Giáo viên

21 tháng 9 2023

a)

$\begin{array}{l}\sin \left( {2x - \frac{\pi }{6}} \right) = - \frac{{\sqrt 3 }}{2}\\ \Leftrightarrow \sin \left( {2x - \frac{\pi }{6}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\end{array}$

$\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{6} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{6} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = - \frac{\pi }{6} + k2\pi \\2x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = \frac{{3\pi }}{4} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}$

b) $\begin{array}{l}\cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \frac{1}{2}\\ \Leftrightarrow \cos \left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right) = \cos \frac{\pi }{3}\end{array}$

$\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{{3x}}{2} + \frac{\pi }{4} = \frac{\pi }{3} + k2\pi \\\frac{{3x}}{2} + \frac{\pi }{4} = \frac{{ - \pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k4\pi }}{3}\\x = \frac{{ - 7\pi }}{{18}} + \frac{{k4\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}$

c)

$\begin{array}{l}\sin 3x - \cos 5x = 0\\ \Leftrightarrow \sin 3x = \cos 5x\\ \Leftrightarrow \cos 5x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} - 3x + k2\pi \\5x = - \left( {\frac{\pi }{2} - 3x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}8x = \frac{\pi }{2} + k2\pi \\2x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{16}} + \frac{{k\pi }}{4}\\x = - \frac{\pi }{4} + k\pi \end{array} \right.\end{array}$

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HQ

Hà Quang Minh

Giáo viên

21 tháng 9 2023

d)

$\begin{array}{l}{\cos ^2}x = \frac{1}{4}\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \frac{1}{2}\\\cos x = - \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = \cos \frac{\pi }{3}\\\cos x = \cos \frac{{2\pi }}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{3} + k2\pi \end{array} \right.\\\left[ \begin{array}{l}x = \frac{{2\pi }}{3} + k2\pi \\x = - \frac{{2\pi }}{3} + k2\pi \end{array} \right.\end{array} \right.\end{array}$

e)

$\begin{array}{l}\sin x - \sqrt 3 \cos x = 0\\ \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{3}.\sin x - \sin \frac{\pi }{3}.\cos x = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 0\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 0\\ \Leftrightarrow x - \frac{\pi }{3} = k\pi ;k \in Z\\ \Leftrightarrow x = \frac{\pi }{3} + k\pi ;k \in Z\end{array}$

f)

$\begin{array}{l}\sin x + \cos x = 0\\ \Leftrightarrow \frac{{\sqrt 2 }}{2}\sin x + \frac{{\sqrt 2 }}{2}\cos x = 0\\ \Leftrightarrow \cos \frac{\pi }{4}.\sin x + \sin \frac{\pi }{4}.\cos x = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin 0\\ \Leftrightarrow x + \frac{\pi }{4} = k\pi ;k \in Z\\ \Leftrightarrow x = - \frac{\pi }{4} + k\pi ;k \in Z\end{array}$

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QN

Quỳnh Nguyễn Thị Ngọc

11 tháng 2 2020

giải phương trình $\sin x\sqrt{1+2\sin x}=\cos2x$ $\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}$ $3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)$ $\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x$ \(\sin2x\sin4x+2\left(3\sin...

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#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

11 tháng 2 2020

a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp

b/ $\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}$

$\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}$

$\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0$

$\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0$

c/ Do $cosx\ne0$, chia 2 vế cho cosx ta được:

$3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)$

Đặt $\sqrt{tanx+1}=t\ge0$

$\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)$

$\Leftrightarrow3t^3-5t^2+3t-10=0$

$\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0$

d/ $\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x$

$\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)$

Đặt $x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi$

$\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa$

$\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0$

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V

Violet

17 tháng 10 2020

1) $sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right).tan^2x-cos^2\frac{x}{2}=0$ 2) $tanx=sin^2x\left(c-\frac{\pi}{2010}\right)+cos^2\left(2x+\frac{\pi}{2010}\right)+sinx.sin\left(3x+\frac{\pi}{1005}\right)$ 3) $1+2cosx\left(sinx-1\right)+\sqrt{2}sinx+4cosx.sin^2\frac{x}{2}=0$ 4) $3cos4x-8cos^6x+2cos4x=3$ 5) $1+sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)$ 6) $sinx.sin4x=\sqrt{2}cos\left(\frac{\pi}{6}-x\right)-4\sqrt{3}cos^2x.sinx.cos2x$ 7)...

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#Toán lớp 11

5

NV

Nguyễn Việt Lâm

Giáo viên

18 tháng 10 2020

Câu 2 bạn coi lại đề

3.

$1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0$

$\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0$

$\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx$

$\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)$

$\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)$

$\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.$

$\Leftrightarrow...$

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NV

Nguyễn Việt Lâm

Giáo viên

18 tháng 10 2020

4.

Bạn coi lại đề, xuất hiện 2 số hạng $cos4x$ ở vế trái nên chắc là bạn ghi nhầm

5.

$\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1$

$\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)$

$\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x$

$\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.$

Xét (1):

$\Leftrightarrow sinx-1-2sinx.cos^2x=0$

$\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0$

$\Leftrightarrow2sin^3x-sinx-1=0$

$\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0$

$\Leftrightarrow...$

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TD

tran duc huy

15 tháng 8 2020

Giải phương trình

1.$sin^2\left(x-\frac{\pi}{6}\right)+sin^2\left(x+\frac{\pi}{6}\right)=5cosx-2$

2.$4sin^4x+12cos^2x=7\\$

3.$sin^4x+cos^{\:4}x=\frac{1}{2}$

4.$sin^4x+cos^4x=cos2x$

#Toán lớp 11

4

NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 8 2020

4.

$\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=cos2x$

$\Leftrightarrow1-\frac{1}{2}sin^22x=cos2x$

$\Leftrightarrow1+1-sin^22x=2cos2x$

$\Leftrightarrow1+cos^22x=2cos2x$

$\Leftrightarrow\left(cos2x-1\right)^2=0$

$\Leftrightarrow cos2x=1$

$\Leftrightarrow2x=k2\pi$

$\Rightarrow x=k\pi$

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NV

Nguyễn Việt Lâm

Giáo viên

15 tháng 8 2020

3.

$\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{1}{2}$

$\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{1}{2}$

$\Leftrightarrow1-sin^22x=0$

$\Leftrightarrow cos^22x=0$

$\Leftrightarrow cos2x=0$

$\Leftrightarrow2x=\frac{\pi}{2}+k\pi$

$\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}$

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QT

Quoc Tran Anh Le

Giáo viên

21 tháng 9 2023

Giải phương trình

a) $\sin \left( {2x + \frac{\pi }{4}} \right) = \sin x$

b) $\sin 2x = \cos 3x$

c) ${\cos ^2}2x = {\cos ^2}\left( {x + \frac{\pi }{6}} \right)$

#Toán lớp 11

1

HQ

Hà Quang Minh

Giáo viên

21 tháng 9 2023

a)

$\sin \left( {2x + \frac{\pi }{4}} \right) = \sin x \Leftrightarrow \left[ \begin{array}{l}2x + \frac{\pi }{4} = x + k2\pi \\2x + \frac{\pi }{4} = \pi - x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\3x = \pi - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{4} + \frac{{k2\pi }}{3}\end{array} \right.;k \in Z$

b)

$\begin{array}{l}\sin 2x = \cos 3x\\ \Leftrightarrow \cos 3x = \cos \left( {\frac{\pi }{2} - 2x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} - 2x + k2\pi \\3x = - \left( {\frac{\pi }{2} - 2x} \right) + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}5x = \frac{\pi }{2} + k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\x = - \frac{\pi }{2} + k2\pi \end{array} \right.\end{array}$

c)

$\begin{array}{l}{\cos ^2}2x = {\cos ^2}\left( {x + \frac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = - \cos \left( {x + \frac{\pi }{6}} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\pi - \left( {x + \frac{\pi }{6}} \right)} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right)\\\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right)\end{array} \right.\end{array}$

Với $\cos 2x = \cos \left( {x + \frac{\pi }{6}} \right) \Leftrightarrow \left[ \begin{array}{l}2x = - \left( {x + \frac{\pi }{6}} \right) + k2\pi \\2x = x + \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{18}} + \frac{{k2\pi }}{3}\\x = \frac{\pi }{6} + k2\pi \end{array} \right.$

Với $\cos 2x = \cos \left( {\frac{{5\pi }}{6} - x} \right) \Leftrightarrow \left[ \begin{array}{l}2x = \frac{{5\pi }}{6} - x + k2\pi \\2x = - \left( {\frac{{5\pi }}{6} - x} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}3x = \frac{{5\pi }}{6} + k2\pi \\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{5\pi }}{{18}} + \frac{{k2\pi }}{3}\\x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right.$

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SB

Sonyeondan Bangtan

18 tháng 10 2021

Giải phương trình sau:
a) $\tan ^2x+4\cos ^2x+7=4\tan x+8\cot x$
b) $6\sin ^2x+2\cos ^2x-2\sqrt{3}\sin 2x=14\sin \left(x-\frac{\pi }{6}\right)$

#Toán lớp 11

0

QN

Quỳnh Nguyễn Thị Ngọc

23 tháng 9 2019

giải phương trình

a, $2\sin\frac{x}{2}\left(\sin\frac{3x}{2}+\cos\frac{3x}{2}\right)=3-4\cos x$

b, $\frac{2\cos^2x+\sqrt{3}\sin2x+3}{2\cos^2x.\sin\left(x+\frac{\pi}{3}\right)}=3\left(\tan^2x+1\right)$

#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

24 tháng 9 2019

a/ $cosx-cos2x+sin2x-sinx=3-4cosx$

$\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0$

$\Leftrightarrow sinx\left(2cosx-1\right)-\left(2cosx-1\right)\left(cosx-2\right)=0$

$\Leftrightarrow\left(2cosx-1\right)\left(sinx-cosx+2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.$

$\Rightarrow x=\pm\frac{\pi}{3}+k2\pi$

b/ ĐKXĐ: $\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.$ $\Rightarrow...$

$\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}$

$\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)$

$\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0$

$\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx-\sqrt{3}\right)+\sqrt{3}sinx\left(2cosx-\sqrt{3}\right)=0$

$\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx+\sqrt{3}sinx-\sqrt{3}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.$ $\Rightarrow...$

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NT

nga thanh

22 tháng 7 2020

Gpt

a, $cos^6x-sin^6x=cos2x$

b, $cos^3x.cos3x-sin^3x.sin3x=\frac{5}{8}$

#Toán lớp 11

1

NV

Nguyễn Việt Lâm

Giáo viên

22 tháng 7 2020

$\left(cos^2x-sin^2x\right)\left(cos^4x+sin^4x+sin^2x.cos^2x\right)=cos2x$

$\Leftrightarrow cos2x\left(cos^4x+sin^4x+sin^2x.cos^2x\right)-cos2x=0$

$\Rightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\cos^4x+sin^4x+sin^2x.cos^2x-1=0\left(1\right)\end{matrix}\right.$

$\left(1\right)\Leftrightarrow\left(sin^2x+cos^2x\right)^2-sin^2x.cos^2x-1=0$

$\Leftrightarrow sin^2x.cos^2x=0$

$\Leftrightarrow sin2x=0\Rightarrow x=\frac{k\pi}{2}$

Tổng hợp lại ta được: $x=\frac{k\pi}{4}$

b/

$\Leftrightarrow\frac{cos3x\left(cos3x+3cosx\right)}{4}+\frac{sin3x\left(3sinx-sin3x\right)}{4}-\frac{5}{8}=0$

$\Leftrightarrow cos^23x+3cosx.cos3x+3sin3x.sin3x-sin^23x-\frac{5}{2}=0$

$\Leftrightarrow\left(cos^23x-sin^23x\right)+3\left(cos3x.cosx+sin3x.sinx\right)-\frac{5}{2}=0$

$\Leftrightarrow cos6x+3cos2x-\frac{5}{2}=0$

$\Leftrightarrow4cos^32x-3cos2x+3cos2x-\frac{5}{2}=0$

$\Leftrightarrow cos^32x=\frac{5}{8}$

$\Rightarrow cos2x=\frac{\sqrt[3]{5}}{2}$

$\Rightarrow x=\pm\frac{1}{2}arccos\left(\frac{\sqrt[3]{5}}{2}\right)+k\pi$

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QT

Quoc Tran Anh Le

Giáo viên

21 tháng 9 2023

Giải các phương trình sau:

a) $\cos \left( {3x - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}$;

b) $2{\sin ^2}x - 1 + \cos 3x = 0$;

c) $\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)$.

#Toán lớp 11

1

HQ

Hà Quang Minh

Giáo viên

21 tháng 9 2023

a) $\cos \left( {3x - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} = - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi + k2\pi }\\{3x = - \frac{\pi }{2} + k2\pi }\end{array}} \right.$

$ \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)$

b) $2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.$

$ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} = - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi + k2\pi }\\{x = - \pi + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)$

c) $\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)$

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