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xuanquynh97
Điểu kiện có nghiệm của phương trình $\sqrt{3}sin2x+cos2x=a$ là $(\sqrt{3})^2+1^2$ \geq a
P
patranopcop
1/[tex] 3sin3x-\sqrt{3}cos9x=2cos3x+4sin^33x[/tex]
2/[tex] sin2x+2sinx-1=4sin^2xcosx+cos2x-2sinxcosx[/tex]
3/[tex]sinx+4cosx-sin2x+2cos2x=1[/tex]
4/[tex] 2sin^3x-cos2x+cosx=0 [/tex]
5/[tex]sin^3x-cos^3x=sinx+cosx[/tex]
6/[tex]8(sin^6x+cos^6x)-3\sqrt{3}sin4x=2[/tex]
giúp mình nha , tks nhiều !
2/[tex] sin2x+2sinx-1=4sin^2xcosx+cos2x-2sinxcosx[/tex]
3/[tex]sinx+4cosx-sin2x+2cos2x=1[/tex]
4/[tex] 2sin^3x-cos2x+cosx=0 [/tex]
5/[tex]sin^3x-cos^3x=sinx+cosx[/tex]
6/[tex]8(sin^6x+cos^6x)-3\sqrt{3}sin4x=2[/tex]
giúp mình nha , tks nhiều !
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xuanquynh97
Bài 2: $sin2x + 2sinx - 1= 4sin^2xcosx + cos2x - 2sinxcos2x$
$<=> sin2x - 4sin^2xcosx + 2sinx - 1 - cos2x + 2sinxcos2x = 0$
$<=> sin2x - 2sinxsin2x + 2sinx - 1 - cos2x + 2sinxcos2x = 0$
$<=> sin2x( 1 - 2sinx ) - ( 1 - 2sinx ) - cos2x( 1 - 2sinx ) = 0$
$<=> ( 1 - 2sinx)( sin2x - 1 - cos2x ) = 0$
$<=> 1 - 2sinx = 0$
hoặc $sin2x - 1 - cos2x = 0$
$<=> sin2x - 4sin^2xcosx + 2sinx - 1 - cos2x + 2sinxcos2x = 0$
$<=> sin2x - 2sinxsin2x + 2sinx - 1 - cos2x + 2sinxcos2x = 0$
$<=> sin2x( 1 - 2sinx ) - ( 1 - 2sinx ) - cos2x( 1 - 2sinx ) = 0$
$<=> ( 1 - 2sinx)( sin2x - 1 - cos2x ) = 0$
$<=> 1 - 2sinx = 0$
hoặc $sin2x - 1 - cos2x = 0$
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trantien.hocmai
$\text{1} \\
PT \leftrightarrow 3\sin 3x-4\sin ^33x-\sqrt{3}\cos 9x=2\cos 3x \\
\leftrightarrow \sin 9x-\sqrt{3}\cos 9x=2\cos 3x \\
\text{2} \\
\sin 2x+2\sin x-1=4\sin ^2x.\cos x+\cos 2x-2\sin x.\cos x \\
\leftrightarrow 2\sin x-1+4\sin x.\cos x+2\sin x-1-4\sin ^2x.\cos x-2\cos ^2x=0 \\
\leftrightarrow 4\sin x.\cos x+2\sin x-4\sin^ 2x.\cos x-2\cos ^2x=0 \\
\text{5} \\
\sin ^3x-\cos ^3x=\sin x+\cos x \\
\leftrightarrow -\sin x(1-\sin ^2x)=\cos x(1+\cos ^2x) \\
\leftrightarrow -\sin x.\cos ^2x=\cos x(1+\cos ^2x) \\
\text{6} \\
8(\sin ^6x+\cos ^6x)-3\sqrt{3}\sin 4x=2 \\
\leftrightarrow 8[(\sin ^2x+\cos ^2x)(\sin ^4x+\cos ^4x-\frac{1}{2} \sin ^22x)]-3\sqrt{3}\sin 4x=2 \\
\leftrightarrow 8(1-\frac{3}{4}\sin ^22x)-3\sqrt{3}\sin 4x=2 \\$
PT \leftrightarrow 3\sin 3x-4\sin ^33x-\sqrt{3}\cos 9x=2\cos 3x \\
\leftrightarrow \sin 9x-\sqrt{3}\cos 9x=2\cos 3x \\
\text{2} \\
\sin 2x+2\sin x-1=4\sin ^2x.\cos x+\cos 2x-2\sin x.\cos x \\
\leftrightarrow 2\sin x-1+4\sin x.\cos x+2\sin x-1-4\sin ^2x.\cos x-2\cos ^2x=0 \\
\leftrightarrow 4\sin x.\cos x+2\sin x-4\sin^ 2x.\cos x-2\cos ^2x=0 \\
\text{5} \\
\sin ^3x-\cos ^3x=\sin x+\cos x \\
\leftrightarrow -\sin x(1-\sin ^2x)=\cos x(1+\cos ^2x) \\
\leftrightarrow -\sin x.\cos ^2x=\cos x(1+\cos ^2x) \\
\text{6} \\
8(\sin ^6x+\cos ^6x)-3\sqrt{3}\sin 4x=2 \\
\leftrightarrow 8[(\sin ^2x+\cos ^2x)(\sin ^4x+\cos ^4x-\frac{1}{2} \sin ^22x)]-3\sqrt{3}\sin 4x=2 \\
\leftrightarrow 8(1-\frac{3}{4}\sin ^22x)-3\sqrt{3}\sin 4x=2 \\$
Last edited by a moderator:
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xuanquynh97
Bài 1, Bài 3 coi lại đề nhá
Bài 4 $2sin^2x.sinx-2cos^2x+cosx+1=0$
\Leftrightarrow $2sinx(1-cos^2x)-2cos^2x+cosx+1=0$
\Leftrightarrow $2sinx(1-cosx)(1+cosx)+(1-cosx)(2cosx+1)=0$
\Leftrightarrow $(1-cosx)(2sinx+2sinxcosx+2cosx+1)=0$
Bài 4 $2sin^2x.sinx-2cos^2x+cosx+1=0$
\Leftrightarrow $2sinx(1-cos^2x)-2cos^2x+cosx+1=0$
\Leftrightarrow $2sinx(1-cosx)(1+cosx)+(1-cosx)(2cosx+1)=0$
\Leftrightarrow $(1-cosx)(2sinx+2sinxcosx+2cosx+1)=0$
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xuanquynh97
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patranopcop
Pt lượng giác
1/[tex] 8cosx=\frac{\sqrt{3}}{sinx}+\frac{1}{cosx} [/tex]
2/ [tex] \sqrt{3}(cos3x+sinx)=sin3x-cosx [/tex]
3/ [tex] 4(sin^4x+cos^4x)+\sqrt{3}sin4x=2 [/tex]
4/ [tex] \frac{cosx-sin2x}{2cos^2x-sinx-1}=\sqrt{3} [/tex]
5/ [tex] tanx-\sqrt{3}=\frac{1}{cosx} [/tex]
6/ [tex] (sinx+cosx)^2+\sqrt{3}cos2x=2 [/tex]
giúp mình nha , tks nhiều !
1/[tex] 8cosx=\frac{\sqrt{3}}{sinx}+\frac{1}{cosx} [/tex]
2/ [tex] \sqrt{3}(cos3x+sinx)=sin3x-cosx [/tex]
3/ [tex] 4(sin^4x+cos^4x)+\sqrt{3}sin4x=2 [/tex]
4/ [tex] \frac{cosx-sin2x}{2cos^2x-sinx-1}=\sqrt{3} [/tex]
5/ [tex] tanx-\sqrt{3}=\frac{1}{cosx} [/tex]
6/ [tex] (sinx+cosx)^2+\sqrt{3}cos2x=2 [/tex]
giúp mình nha , tks nhiều !
M
mua_sao_bang_98
2. $\sqrt{3} (cos3x+sinx)=sin3x-cosx$
\Leftrightarrow $\sqrt{3}cos3x-sin3x=-(cosx+\sqrt{3}sinx)$
\Leftrightarrow $-sin(3x-\frac{\pi}{3})=-sin(x+\frac{\pi}{6})$
\Leftrightarrow $sin(3x-\frac{\pi}{3})= sin(x+\frac{\pi}{6})$
dễ rồi nhé! ^^
\Leftrightarrow $\sqrt{3}cos3x-sin3x=-(cosx+\sqrt{3}sinx)$
\Leftrightarrow $-sin(3x-\frac{\pi}{3})=-sin(x+\frac{\pi}{6})$
\Leftrightarrow $sin(3x-\frac{\pi}{3})= sin(x+\frac{\pi}{6})$
dễ rồi nhé! ^^
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mua_sao_bang_98
3. $4(sin^4x+ cos^4x)+\sqrt{3}sin4x=2$
\Leftrightarrow $4[(sin^2x+cos^2x)^2-2sin^2xcos^2x] + \sqrt{3}sin4x=2$
\Leftrightarrow $4(1-2sin^2xcos^2x) + 2\sqrt{3}sin2xcos2x=2$
\Leftrightarrow $4-2sin^22x+2\sqrt{3}sin2xcos2x=2$
\Leftrightarrow $2-2sin^22x+\sqrt{3}sin4x=0$
\Leftrightarrow $sin4x+\sqrt{3}sin4x=-1$
\Leftrightarrow $sin(4x+\frac{\pi}{3})=-1$
dễ rồi ha!
\Leftrightarrow $4[(sin^2x+cos^2x)^2-2sin^2xcos^2x] + \sqrt{3}sin4x=2$
\Leftrightarrow $4(1-2sin^2xcos^2x) + 2\sqrt{3}sin2xcos2x=2$
\Leftrightarrow $4-2sin^22x+2\sqrt{3}sin2xcos2x=2$
\Leftrightarrow $2-2sin^22x+\sqrt{3}sin4x=0$
\Leftrightarrow $sin4x+\sqrt{3}sin4x=-1$
\Leftrightarrow $sin(4x+\frac{\pi}{3})=-1$
dễ rồi ha!
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mua_sao_bang_98
4. $\frac{cosx-sin2x}{2cos^2x-sinx-1}=\sqrt{3}$
ĐK: $2cos^2x-sinx-1 \neq 0$ \Leftrightarrow $1-2sin^2x-sinx \neq 0$ \Rightarrow giải ra nhé! ^^
pt \Leftrightarrow $cosx-sin2x= \sqrt{3}(cos2x-sinx)$
\Leftrightarrow $cosx+\sqrt{3}sinx=sin2x+\sqrt{3}cos2x$
\Leftrightarrow $sin(x+\frac{\pi}{6})=sin(2x+\frac{\pi}{3})$
Đến đây tự làm được rồi nhỉ? ^^
ĐK: $2cos^2x-sinx-1 \neq 0$ \Leftrightarrow $1-2sin^2x-sinx \neq 0$ \Rightarrow giải ra nhé! ^^
pt \Leftrightarrow $cosx-sin2x= \sqrt{3}(cos2x-sinx)$
\Leftrightarrow $cosx+\sqrt{3}sinx=sin2x+\sqrt{3}cos2x$
\Leftrightarrow $sin(x+\frac{\pi}{6})=sin(2x+\frac{\pi}{3})$
Đến đây tự làm được rồi nhỉ? ^^
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mua_sao_bang_98
5. $tanx-\sqrt{3}=\frac{1}{cosx}$
ĐK: $cosx \neq 0$ \Leftrightarrow $x \neq \frac{\pi}{2}+ k\pi$
pt \Leftrightarrow $\frac{sinx}{cosx}-\sqrt{3}=\frac{1}{cosx}$
\Leftrightarrow $sinx-\sqrt{3}cosx=1$
\Leftrightarrow $sin(x-\frac{\pi}{3})=\frac{1}{2}=sin(\frac{\pi}{3})$
ĐK: $cosx \neq 0$ \Leftrightarrow $x \neq \frac{\pi}{2}+ k\pi$
pt \Leftrightarrow $\frac{sinx}{cosx}-\sqrt{3}=\frac{1}{cosx}$
\Leftrightarrow $sinx-\sqrt{3}cosx=1$
\Leftrightarrow $sin(x-\frac{\pi}{3})=\frac{1}{2}=sin(\frac{\pi}{3})$
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patranopcop
giải pt lượng giác với sinx ; cosx
1/[tex] cos^2x-\sqrt{3}sin2x=1+sin^2 [/tex]
2/[tex] cos(\frac{\pi}{2}+2x)-\sqrt{3}cos(\pi-2x)=1 [/tex]
3/[tex] 2sin2xcos2x+\sqrt{3}cos4x=\sqrt{2} [/tex]
4/[tex] 8cosx=\frac{\sqrt{3}}{sinx}+\frac{1}{cosx} [/tex]
5/[tex] 3sin3x-\sqrt{3}cos9x=2cos3x+4sin^33x[/tex]
6/[tex] sinx+4cosx-sin2x+2cos2x=1[/tex]
giúp mình nha , tks nhiều !
1/[tex] cos^2x-\sqrt{3}sin2x=1+sin^2 [/tex]
2/[tex] cos(\frac{\pi}{2}+2x)-\sqrt{3}cos(\pi-2x)=1 [/tex]
3/[tex] 2sin2xcos2x+\sqrt{3}cos4x=\sqrt{2} [/tex]
4/[tex] 8cosx=\frac{\sqrt{3}}{sinx}+\frac{1}{cosx} [/tex]
5/[tex] 3sin3x-\sqrt{3}cos9x=2cos3x+4sin^33x[/tex]
6/[tex] sinx+4cosx-sin2x+2cos2x=1[/tex]
giúp mình nha , tks nhiều !
C
chaizo1234567
câu 1
Ta có
$4(2cosx-1)-\frac{2sinx-\sqrt{3}}{sinx}-\frac{2cosx-1}{cosx}=0$
Ta luôn có
$2sinx-\sqrt{3}=2cosx-1=0$ vơi mọi x
\Rightarrow$(2cosx-1)(4-\frac{1}{sinx}-\frac{1}{cosx})=0$
Nếu có gì sai mong các bạn thông cảm
.................
Ta có
$4(2cosx-1)-\frac{2sinx-\sqrt{3}}{sinx}-\frac{2cosx-1}{cosx}=0$
Ta luôn có
$2sinx-\sqrt{3}=2cosx-1=0$ vơi mọi x
\Rightarrow$(2cosx-1)(4-\frac{1}{sinx}-\frac{1}{cosx})=0$
Nếu có gì sai mong các bạn thông cảm
.................
Last edited by a moderator:
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mua_sao_bang_98
1. $cos^2x-\sqrt{3}sin2x=1+sin^2x$
\Leftrightarrow $cos^2x-\sqrt{3}sin2x=sin^2x+cos^2x+sin^2x$
\Leftrightarrow $2sin^2x+2\sqrt{3}sinxcosx=0$
\Leftrightarrow $sinx(sinx+\sqrt{3}cosx)=0$
\Leftrightarrow $\left[\begin{matrix} sinx= 0 \\ sin(x+\frac{\pi}{3})=0 \end{matrix}\right.$
\Leftrightarrow$ \left[\begin{matrix} x=k\pi \\ x=\frac{-\pi}{3}+k\pi \end{matrix}\right.$
\Leftrightarrow $cos^2x-\sqrt{3}sin2x=sin^2x+cos^2x+sin^2x$
\Leftrightarrow $2sin^2x+2\sqrt{3}sinxcosx=0$
\Leftrightarrow $sinx(sinx+\sqrt{3}cosx)=0$
\Leftrightarrow $\left[\begin{matrix} sinx= 0 \\ sin(x+\frac{\pi}{3})=0 \end{matrix}\right.$
\Leftrightarrow$ \left[\begin{matrix} x=k\pi \\ x=\frac{-\pi}{3}+k\pi \end{matrix}\right.$
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mua_sao_bang_98
2. $cos(\frac{\pi}{2}+2x)-\sqrt{3}cos(\pi-2x)=1$
\Leftrightarrow $-sin2x+\sqrt{3}cos2x=1$
\Leftrightarrow $sin(\frac{\pi}{3}-2x)=\frac{1}{2}=sin(\frac{\pi}{6})$
\Leftrightarrow $\left[\begin{matrix} \frac{\pi}{3}-2x=\frac{\pi}{6}+k2\pi \\ \frac{\pi}{3}-2x =\frac{5\pi}{6}+k2\pi \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} x=\frac{\pi}{12}+k\pi \\ x=\frac{-\pi}{4}+k\pi \end{matrix}\right.$
\Leftrightarrow $-sin2x+\sqrt{3}cos2x=1$
\Leftrightarrow $sin(\frac{\pi}{3}-2x)=\frac{1}{2}=sin(\frac{\pi}{6})$
\Leftrightarrow $\left[\begin{matrix} \frac{\pi}{3}-2x=\frac{\pi}{6}+k2\pi \\ \frac{\pi}{3}-2x =\frac{5\pi}{6}+k2\pi \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} x=\frac{\pi}{12}+k\pi \\ x=\frac{-\pi}{4}+k\pi \end{matrix}\right.$
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mua_sao_bang_98
3. $2sin2xcos2x+\sqrt{3}cos4x=\sqrt{2}$
\Leftrightarrow $sin4x+\sqrt{3}cos4x=\sqrt{2}$
\Leftrightarrow $sin(4x+\frac{\pi}{3})=\frac{1}{\sqrt{2}}=sin\frac{\pi}{4}$
\Leftrightarrow $\left[\begin{matrix} 4x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi \\ 4x+\frac{\pi}{3}=\frac{3\pi}{4}+k2\pi \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} x=\frac{-\pi}{48}+k\frac{\pi}{2} \\ x=\frac{5\pi}{48}+k\frac{\pi}{2} \end{matrix}\right.$
\Leftrightarrow $sin4x+\sqrt{3}cos4x=\sqrt{2}$
\Leftrightarrow $sin(4x+\frac{\pi}{3})=\frac{1}{\sqrt{2}}=sin\frac{\pi}{4}$
\Leftrightarrow $\left[\begin{matrix} 4x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi \\ 4x+\frac{\pi}{3}=\frac{3\pi}{4}+k2\pi \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} x=\frac{-\pi}{48}+k\frac{\pi}{2} \\ x=\frac{5\pi}{48}+k\frac{\pi}{2} \end{matrix}\right.$
Last edited by a moderator:
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mua_sao_bang_98
$sin9x-\sqrt{3}cos9x=2cos3x$
\Leftrightarrow $\frac{1}{2}sin9x-\frac{\sqrt{3}}{2}cos9x=cos3x$
\Leftrightarrow $cos(9x+\frac{\pi}{6})+cos3x=0$
\Leftrightarrow $sin(6x+\frac{\pi}{12})sin(3x+\frac{\pi}{12})=0$
\Leftrightarrow $\left[\begin{matrix} sin(6x+\frac{\pi}{12})=0 \\ sin(3x+\frac{\pi}{12})=0 \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} 6x+\frac{\pi}{12}=k\pi \\ 3x+\frac{\pi}{12}=k\pi \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} x=-\frac{\pi}{72}+k\frac{\pi}{6} \\ x=-\frac{\pi}{36}+k\frac{\pi}{3} \end{matrix}\right.$
\Leftrightarrow $\frac{1}{2}sin9x-\frac{\sqrt{3}}{2}cos9x=cos3x$
\Leftrightarrow $cos(9x+\frac{\pi}{6})+cos3x=0$
\Leftrightarrow $sin(6x+\frac{\pi}{12})sin(3x+\frac{\pi}{12})=0$
\Leftrightarrow $\left[\begin{matrix} sin(6x+\frac{\pi}{12})=0 \\ sin(3x+\frac{\pi}{12})=0 \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} 6x+\frac{\pi}{12}=k\pi \\ 3x+\frac{\pi}{12}=k\pi \end{matrix}\right.$
\Leftrightarrow $\left[\begin{matrix} x=-\frac{\pi}{72}+k\frac{\pi}{6} \\ x=-\frac{\pi}{36}+k\frac{\pi}{3} \end{matrix}\right.$
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xuanquynh97
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mua_sao_bang_98
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xuanquynh97
Sai mình làm lại
a.ĐK:
[TEX] sin2x \neq 0 \Leftrightarrow x \neq \frac{k \pi}{2}[/TEX]
PT[TEX]\Leftrightarrow 8sin^2xcosx= \sqrt{3}sinx+cosx(1)[/TEX]
[TEX] \Leftrightarrow 8cosx(1-cos^2x)= \sqrt{3}sinx+cosx[/TEX]
[TEX]\Leftrightarrow 8cosx-8cos^3x= \sqrt{3}sinx+cosx[/TEX]
[TEX]\Leftrightarrow 2(3cosx-4cos^3x)= \sqrt{3}sinx-cosx[/TEX]
[TEX]\Leftrightarrow -2cos3x=-2cos(x+ \frac{ \pi}{3})[/TEX]