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#algebra , #trigonometry, #numbersystem
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The latest Messages 20
2021-12-26 15:19:19
sin⁴x-8sin²x+4=0
(sin²x-2sinx-2)(sin²x+2sinx-2)=0
sin²x-2sinx=2
251 views dɹɐʇᴉʞ , 12:19
2021-12-26 15:03:04
Let sintheta = x
x² + 4/x² = 8
x - 2/x = 2
x² - 2x = 2 ans
265 viewsBunny, 12:03
2021-12-26 14:25:59
306 views dɹɐʇᴉʞ , 11:25
2021-12-26 14:25:47
296 views dɹɐʇᴉʞ , 11:25
2021-12-25 11:03:59
equation 1×2+equation 2
√64 = 8
162 viewsBunny, 08:03
2021-12-25 11:00:31
172 viewsBunny, 08:00
2021-12-25 10:58:49
y=(x-5)/2
(X-5)/2=(2x²+18x-3x-27)
x-5=2(2x²+15x-27)
x-5=4x²+30x-54
4x²+29x-49=0
Discriminant=529+16×49>0
So this equation has two distinct real roots.
So for two distinct values of x there will be two distinct values of y.
So 2 pairs of points will satisfy both equation. Hence 2 solutions
172 viewsBunny, 07:58
2021-12-25 10:55:29
140 viewsBunny, 07:55
2021-12-25 10:52:51
x+1 = y²+1-2y
y+1 = x²+1-2x
Add
x²+y² = 3(x+y)
Substract
x-y = (y²-x²)-2(y-x)
x+y= 1
Req: (3/3)²⁰²⁰ = 1 ans
139 viewsBunny, 07:52
2021-12-25 10:49:31
(x+1)==(y+1)²_______(1)
(y+1)==(x–1)²_______(2)
Equation (1) – (2)
°°°°°°°°°°°°°°°°°°°°°°
(x–y)==(y²+1–2y)-(x²+1-2x)
(x-y)==y² - x² - 2y +2x
x-y== (y+x-2)(y-x)
(y+x-2)== -1
x+y== 1
Equation (1)
x+1==(y-1)²
x+1== x²
x²-x-1==0 & y²-y-1==0
x²+y²-(x+y)-2==0
x²+y²-1-2=0
x²+y²==3
atq......
[(x²+y²)/3]^2020
[(3/3)]^2020
1
166 viewsMr᭄jituᴮᴼˢˢ࿐ Ǥuקtα, 07:49
A
