PTLG 11 ??? Gấp

[tiendungyamaha]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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ĐĂNG BÀI NGAY để cùng trao đổi với các thành viên siêu nhiệt tình & dễ thương trên diễn đàn.

1. [TEX]12Cosx +5Sinx +\frac{5}{12Cosx + 5Sinx +14} +8 =0[/TEX]
2. [TEX]Sinx + 2sin3x = - sin5x[/TEX]
3. [TEX]Cos5x.Cosx = Cos4x[/TEX]
4. [TEX]Sinx.Sin2x.Sin3x = \frac{1}{4}Sin4x[/TEX]
5. [TEX]Sin^4x + Cos^4x = -\frac{1}{2}Cos^22x[/TEX]
6. [TEX]Sin^6x + Cos^6x = 4Cos^22x[/TEX]
7. [TEX] -\frac{1}{4} + Sin^2x = Cos^4x [/TEX]
8. [TEX]Cos^2x = 3Sin2x + 3[/TEX]
9. [TEX]Cotx - Cot2x = Tanx +1[/TEX]
10. [TEX]8Cot^4x - 4Cos2x +sin4x - 4 =0[/TEX]
11. [TEX]Sin^6x + Cos^6x + \frac{1}{4}Sin4x = 0[/TEX]
12. [TEX]Cos^4x +Sin^4x + Cos(x-\frac{\pi}{4}).Sin(3x-\frac{\pi}{4}) - \frac{3}{2}=0[/TEX]
13. [TEX]1 +3Tanx = 2Sin2x [/TEX]
14. [TEX] -2 + 4Cos^2x - Cos3x = 6Cosx + 2Cos2x [/TEX]
15. [TEX]Sin^2x + Sin^23x - 3Cos^22x = 0[/TEX]

 

[chaizo1234567]

PT
\Leftrightarrow4sinx.sin2x.sin3x=sin4x
\Leftrightarrow2sin2x(cos2x-cos4x)=sin4x
\Leftrightarrow2sin2x.cos4x=0
...........................

PT
\Leftrightarrow$2(cosx^2+sinx^2)-4sinx^2.cosx^2=-cos2x^2$
\Leftrightarrowcos4x=-2
vậy phương trình vô nghiệm....

áp dụng công thức
cotx-tagx=2cot2x
PT
\Leftrightarrowcotx-3tagx=2
\Leftrightarrow$cosx^2-3sinx^2-2sinx.cosx=0$
Đến đây đặt điều kiện và giải là xong.....

 

[quynhsieunhan]

1, Đặt $12cosx + 5sinx + 14 = t$
PT trở thành: $t + \frac{5}{t} + 6 = 0$
\Rightarrow $t^2 - 6y + 5 = 0$

 

[quynhsieunhan]

6, $sin^6x + cos^6x = 4cos^22x$
\Leftrightarrow $1 - 3sin^2xcos^2x = 4 - 4sin^22x$
\Leftrightarrow $\frac{4 - 3sin^22x}{4} = 4 - 4sin^22x$
\Leftrightarrow $4 - 3sin^22x = 16 - 16sin^22x$
\Leftrightarrow $13sin^22x = 12$

8, $cos^2x = 3sin2x + 3$
\Leftrightarrow $1 + cos2x = 6sin2x + 6$
\Leftrightarrow $6sin2x - cos2x = -5$

9, $cotx - cot2x = tanx + 1$
\Leftrightarrow $\frac{cosx}{sinx} - \frac{cos2x}{sin2x} = \frac{sinx}{cosx} + 1$
\Leftrightarrow $\frac{1}{sin2x} = \frac{sinx}{cosx} + 1$
\Rightarrow $1 = 2sin^2x + sin2x$
\Leftrightarrow $sin2x = cos2x$

11, $sin^6x + cos^6x + \frac{1}{4}sin4x = 0$
\Leftrightarrow $\frac{4 - 3sin^22x}{4} + \frac{1}{4}sin4x = 0$
\Leftrightarrow $4 - \frac{3}{2}(1 - cos4x) + sin4x = 0$
\Leftrightarrow $\frac{3}{2}cos4x + sin4x + \frac{5}{2} = 0$

 

[mua_sao_bang_98]

1. $12cosx+5sinx+\frac{5}{12cosx+5sinx+14}+8=0$

Đặt 12cosx+5sinx=t (t#-14) , ta có:

pt \Leftrightarrow $t+\frac{5}{t+14}+8=0$

\Leftrightarrow $t^2+14t+5+8t+112=0$

\Leftrightarrow $\left[\begin{matrix} 12cosx+5sinx=-9 \\ 12cosx+5sinx=-13 \end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} \frac{12}{13}cosx+\frac{5}{13}sinx=\frac{-9}{13} \\ \frac{12}{13}cosx+\frac{5}{13}sinx=-1 \end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} cos(x-a)=\frac{-9}{13} \\ cos(x-a)=-1 \end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} x=a+arccos(\frac{-9}{13}) +k2\pi \\ x=a+\pi+k2\pi \end{matrix}\right.$

2. $sinx+2sin3x=-sin5x$

\Rightarrow $ sinx+sin5x+2sin3x=0$

\Leftrightarrow $2sin3x.cos2x+2sin3x=0$

\Leftrightarrow $sin3x(cos2x+1)=0$

\Leftrightarrow $\left[\begin{matrix} sin3x=0 \\ cos2x=-1 \end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} x=\frac{k\pi}{6} \\ x=\frac{-\pi}{4}+k\pi \end{matrix}\right.$

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3. $cos5x.cosx=cos4x$

\Leftrightarrow $cos4x+cos6x=2cos4x$

\Leftrightarrow $cos4x-cos6x=0$

\Leftrightarrow $2sin5x.sinx=0$

\Leftrightarrow $\left[\begin{matrix} sin5x=0 \\ sinx=0 \end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} x=\frac{k\pi}{10} \\ x=\frac{k\pi}{2}\end{matrix}\right.$

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4.$sinx.sin2x.sin3x=\frac{1}{4}sin4x$

\Leftrightarrow $sinx.sin2x.sin3x=\frac{1}{4}.2sin2xcos2x$

\Leftrightarrow $2sinxsin3x=cos2x$

\Leftrightarrow $2sin^2x(3-4sin^2x)=1-2sin^2x$

\Leftrightarrow $6sin^2x-8sin^4x-1+2sin^2x=0$

\Leftrightarrow $8sin^4x-8sin^2x+1=0$

\Leftrightarrow $\left[\begin{matrix} sin^2x=\frac{2+\sqrt{2}}{4} \\ sin^2x=\frac{2-\sqrt{2}}{4}\end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} sinx=\pm \sqrt{\frac{2+\sqrt{2}}{4}} \\ sinx=\pm \sqrt{\frac{2-\sqrt{2}}{4}}\end{matrix}\right.$

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

~~> Bài này không biết làm đúng không mà sao số to gớm.

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5. $sin^4x+cos^4x=-\frac{1}{2}cos^22x$

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

\Leftrightarrow $2-sin^22x=-cos^22x$

\Leftrightarrow $sin^22x-cos^22x=2$

\Leftrightarrow $cos4x=-2$ \Rightarrow pt vô nghiệm.

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6. $sin^6x+cos^6=4cos^22x$

\Leftrightarrow $sin^4x+cos^4x-sin^2xcos^2x=4cos^22x$

\Leftrightarrow $1-3sin^2xcos^2x=4cos^22x$

\Leftrightarrow $4-3sin^22x=4cos^22x$

\Leftrightarrow $3sin^22x+16cos^22x=16$

\Leftrightarrow $3-3cos^22x+16cos^22x=16$

\Leftrightarrow $cos^22x=1$

\Leftrightarrow $\left[\begin{matrix}cos2x=1 \\ cos2x=-1 \end{matrix}\right.$

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

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7. $\frac{-1}{4}+sin^2x=cos^4x$

\Leftrightarrow $-1+4sin^2x-4cos^4x=0$

\Leftrightarrow $-1+4sin^2x-4(1-sin^2x)^2=0$

\Leftrightarrow $-1+4sin^2x-4+8sin^2x-4sin^4x=0$

\Leftrightarrow $4sin^4x-12sin^2x+5=0$

\Leftrightarrow $\left[\begin{matrix} sin^2x=\frac{5}{2} (Vô nghiệm) \\ sin^2x=\frac{1}{2} \end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} x=\frac{\pi}{4}+k2\pi \\ x =\frac{3\pi}{4}+k2\pi \\ x=-\frac{\pi}{4}+k2\pi \\ x=\frac{5\pi}{4}+k2\pi \end{matrix}\right.$

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8. $cos^2x=3sin2x+3$

\Leftrightarrow $\frac{1+cos2x}{2}=3sin2x+3$

\Leftrightarrow $1+cos2x=6sin2x+6$

\Leftrightarrow $cos2x-6sin2x=5$

\Leftrightarrow $\frac{1}{\sqrt{37}}cos2x-\frac{6}{\sqrt{37}}sin2x=\frac{5}{\sqrt{37}}$

Đặt $\frac{1}{\sqrt{27}}=cost$

\Leftrightarrow $cos(t+2x)=\frac{5}{\sqrt{27}}$

\Leftrightarrow $\left[\begin{matrix} t+2x=arccos(\frac{5}{\sqrt{27}}) +k2\pi \\ t+2x= - arccos(\frac{5}{\sqrt{27}})\end{matrix}\right.$

\Leftrightarrow $\left[\begin{matrix} x=\frac{-1+arccos(\frac{5}{\sqrt{27}})}{2} +k\pi \\ x=\frac{-1-arccos(\frac{5}{\sqrt{27}})}{2} +k\pi \end{matrix}\right.$

 

[mua_sao_bang_98]

9. $cotx-cot2x=tanx+1$

ĐK: $\left\{\begin{matrix} sinx \neq 0 \\ sin2x \neq 0 \\ cosx \neq 0 \end{matrix}\right.$

\Rightarrow ĐK: $ x \neq k\pi $

pt \Leftrightarrow $\frac{cosx}{sinx}-\frac{cos2x}{sin2x}=\frac{sinx}{cosx}+1$

\Leftrightarrow $2cos^2x-cos2x=2sin^2x+sin2x$

\Leftrightarrow $2(cos^2x-sin^2x)=sin2x+cos2x$

\Leftrightarrow $2cos2x=sin2x+cos2x$

\Leftrightarrow $cos2x-sin2x=0$

\Leftrightarrow $cos(2x+\frac{\pi}{4})=0$

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

\Leftrightarrow $x=\frac{\pi}{8}+\frac{k\pi}{2}$

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10. $8cot^4x-4cos2x+sin4x-4=0$

ĐK: $sinx \neq 0 $ \Leftrightarrow $x\neq k\pi$

pt \Leftrightarrow $8cot^4x-4(cos2x+1)+sin4x=0$

\Leftrightarrow $8\frac{cos^4x}{sin^4x}-8cos^2x+sin4x=0$

\Leftrightarrow $8cos^4x-8cos^2x.sin^4x+sin4x.sin^4x=0$

~~ Từ chỗ này may ra là ra nhưng mà tớ chưa nghĩ ra cách khai triển tiếp! cứ để đây đã.
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