PT bậc n (n>=2)

[teddyhandsome]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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1. cos4x + 12sin^2(x) - 1 = 0..
2. 4sin^4(x) + 12cos^2(x) = 7.
3. cos2x + 4sin^4(x) = 8cos^6(x)
4. 3 - 4cos^2(x) = (2sinx +1)sinx
5. sin^4(x) + cos^4(x) + (1/2).sin^2(2x) = 2 + cos2x
6. cosx - căn2.sin(x/2) = 1.
7. cos^2(6x).cos4x = cos^2(2x)

 

[newstarinsky]

1. cos4x + 12sin^2(x) - 1 = 0..
2. 4sin^4(x) + 12cos^2(x) = 7.
3. cos2x + 4sin^4(x) = 8cos^6(x)
4. 3 - 4cos^2(x) = (2sinx +1)sinx
5. sin^4(x) + cos^4(x) + (1/2).sin^2(2x) = 2 + cos2x
6. cosx - căn2.sin(x/2) = 1.
7. cos^2(6x).cos4x = cos^2(2x)

$1) 2cos^22x-1+6(1-cos2x)-1=0\\
\Leftrightarrow 2cos^22x-6cos2x+4=0\\

2) 4(1-cos^2x)^2+12cos^2x=7\\
\Leftrightarrow 4cos^4x+4cos^2x-3=0\\

3) 2cos^2x-1+4(1-cos^2x)^2=8cos^6x\\
\Leftrightarrow 8cos^6x-4cos^4x+6cos^2x-3=0\\
Đặt cos^2x=t \\

4) 3-4(1-sin^2x)=2sin^2x+sinx\\
\Leftrightarrow 2sin^2x-sinx-1=0\\

5)(sin^2x+Cos^2x)^2-2sin^2x.cos^2x+2sin^2x.cos^2x=2+cos2x\\
\Leftrightarrow 1=2+cos2x\\
\Leftrightarrow cos2x=-1\\

6)1-2sin^2\dfrac{x}{2}-\sqrt{2}sin\dfrac{x}{2}=1\\
\Leftrightarrow 2sin^2\dfrac{x}{2}+\sqrt{2}sin\dfrac{x}{2}=0\\

7)(1+cos12x)cos4x=1+cos4x\\
\Leftrightarrow cos12x.cos4x=1\\
\Leftrightarrow (4cos^34x-3cos4x)cos4x=1\\
\Leftrightarrow 4cos^44x-3cos^24x-1=0$

 

[teddyhandsome]

8. 5cosx - 2 = 3(1-cosx).cot^2(x)
9. 3(tanx + cotx) = 2(2 + sin2x)
10. sin^5(x) + cos^6(x) = 1
11. sin^4(x) + cos^3(x) = 1
12. (sin^4(2x) + cos^4(2x))/tan(pi/4 - x).tan(pi/4 + x) = cos^4(4x)
13. 2sinx - cosx + (1 +sin2x)/sinx + cosx = cos2x - 2
14. 4cosx - sinx + (1 - sin2x)/căn2.sin(x - pi/4) = 2 + cos2x
15. 1/sin^2(x) - (1 + căn3)cotx + căn3 -1 = 0
16. tan^3(x) + 1/cos^2(x) = 4 + 3cot(pi/2 -x)
17. ((1 + sinx + cos2x).sin(x + pi/4))/1 + tanx = cosx/căn2
18. 4cos(5x/2).cos(3x/2) + 2(8sinx -1).cosx = 5
19. (sin2x + căn3.cos2x)^2 - 3 = cos(pi/6 - 2x)
20. (cos2x + 3cot2x + sin4x)/cot2x - cos2x = 2

 

[newstarinsky]

8. 5cosx - 2 = 3(1-cosx).cot^2(x)
9. 3(tanx + cotx) = 2(2 + sin2x)
10. sin^5(x) + cos^6(x) = 1
11. sin^4(x) + cos^3(x) = 1
12. (sin^4(2x) + cos^4(2x))/tan(pi/4 - x).tan(pi/4 + x) = cos^4(4x)
13. 2sinx - cosx + (1 +sin2x)/sinx + cosx = cos2x - 2

$8)5cosx-2=3(1-cosx)\dfrac{cos^2x}{sin^2x}\\
\Leftrightarrow (1-cosx)(1+cosx)(5cosx-2)=3(1-cosx)cos^2x\\
\Leftrightarrow (1-cosx)(2cos^2x+3cosx-2)=0\\

9)3(\dfrac{sin^2x+cos^2x}{sinx.cosx})=2(2+sin2x)\\
\Leftrightarrow 6=2sin2x(2+sin2x)\\
\Leftrightarrow sin^22x+2sin2x-3=0\\

10)sin^3x.sin^2x=1-cos^6x\\
\Leftrightarrow sin^3x.sin^2x=(1-cos^2x)(1+cos^2x+cos^4x)\\
\Leftrightarrow sin^2x(sin^3x-1-cos^2x-cos^4x)=0$
ta có $1+cos^2x+cos^4x>1$ nên$ sin^3x-1-cos^2x-cos^4x<0$ nên PT trong ngoặc vô nghiệm

$11)cos^3x=1-sin^4x\\
\Leftrightarrow cos^3x=(1-sin^2x)(1+sin^2x)\\
\Leftrightarrow cos^3x=cos^2x(1+sin^2x)\\
\Leftrightarrow cos^2x(1+sin^2x-cosx)=0\\
\Leftrightarrow cos^2x(2-cos^2x-cosx)=0\\

12)cos(\dfrac{\pi}{4}-x).cos(\dfrac{\pi}{4}+x).[(sin^22x+cos^22x)^2-2sin^22x.cos^22x]=sin(\dfrac{\pi}{4}-x)sin(\dfrac{\pi}{4}+x).cos^44x\\
\Leftrightarrow cos2x(1-\dfrac{1}{2}sin^24x)=cos2x.cos^44x\\
\Leftrightarrow cos2x(2-sin^24x-2cos^44x)=0\\
\Leftrightarrow cos2x[2(1-cos^24x)(1+cos^24x)-sin^24x]=0\\
\Leftrightarrow cos2x[2sin^24x(2-sin^24x)-sin^24x]=0\\
\Leftrightarrow cos2x(-2sin^44x+3sin^24x)=0\\

13)2sinx-cosx+\dfrac{(sinx+cosx)^2}{sinx+cosx}=cos2x-2\\
\Leftrightarrow 2sinx-cosx+sinx+cosx=1-2sin^2x-2\\
\Leftrightarrow 3sinx=-1-2sin^2x\\
\Leftrightarrow 2sin^2x+3sinx+1=0\\$

 

[newstarinsky]

14. 4cosx - sinx + (1 - sin2x)/căn2.sin(x - pi/4) = 2 + cos2x
15. 1/sin^2(x) - (1 + căn3)cotx + căn3 -1 = 0
16. tan^3(x) + 1/cos^2(x) = 4 + 3cot(pi/2 -x)
17. ((1 + sinx + cos2x).sin(x + pi/4))/1 + tanx = cosx/căn2
18. 4cos(5x/2).cos(3x/2) + 2(8sinx -1).cosx = 5
19. (sin2x + căn3.cos2x)^2 - 3 = cos(pi/6 - 2x)
20. (cos2x + 3cot2x + sin4x)/cot2x - cos2x = 2

$14)4cosx-sinx+\dfrac{(sinx-cosx)^2}{sinx-cosx}=co2x+2\\
\Leftrightarrow 3cosx=2cos^2x+1$

$15)1+cot^2x-(1+\sqrt{3})cotx+\sqrt{3}-1=0\\
\Leftrightarrow cot^2x-(1+\sqrt{3})cotx+\sqrt{3}=0\\

16)tan^3x+1+tan^2x=4+3tanx\\
\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\\
\Leftrightarrow (tanx+1)(tan^2x-3)=0$

$17) \dfrac{[cosx(1+sinx+cos2x)(sinx+cosx)]}{\sqrt{2}(sinx+cosx)}=\dfrac{cosx}{\sqrt{2}}\\
\Leftrightarrow cosx(2+sinx-2sin^2x)=0$

$18)2cos4x+2cosx+2(8sinx-1)cosx=5\\
\Leftrightarrow 2(1-2sin^22x)+8sin2x=5\\
\Leftrightarrow -4sin^22x+8sin2x-3=0$

$19)4cos^2(\dfrac{\pi}{6}-2x)-3=cos(\dfrac{\pi}{6}-2x)\\

20)\dfrac{cos2x.sin2x+3cos2x+2sin2x.cos2x}{cos2x-cos2x.sin2x}=2\\
\Leftrightarrow 3sin2x+3=2(1-sin2x)\\
\Leftrightarrow 5sin2x=-1\\ $