Sin³ x-sin x=cos ² x
we know that:
sin²x + cos²x=1 ⇒cos²x=1-sin²x
Therefore:
sin³x-sin x=1-sin²x
sin³x+sin²x-sin x-1=0
sin³x=z
z³+z²-z-1=0
we divide by Ruffini method:
1 1 -1 -1
1 1 2 1 z=1
-------------------------------------
1 2 1 0
-1 -1 -1 z=-1
--------------------------------------
1 1 0 z=-1
Therefore; the solutions are z=-1 and z=1
The solutions are:
if z=-1, then
sin x=-1 ⇒x= arcsin -1=π+2kπ (180º+360ºK) K∈Z
if z=1, then
sin x=1 ⇒ x=arcsin 1=π/2 + 2kπ (90º+360ºK) k∈Z
π/2 + 2kπ U π+2Kπ=π/2+kπ k∈Z ≈(90º+180ºK)
Answer: π/2 + Kπ or 90º+180ºK K∈Z
Z=...-3,-2,-1,0,1,2,3,4....
We have to determine the complete factored form of the given polynomial
.
Let x= -1 in the given polynomial.
So, 
So, by factor theorem
(x+1) is a factor of the given polynomial.
So, dividing the given polynomial by (x+1), we get quotient as 
.
So, 
= (x+1).
= 
=![(x+1)[ 2x(3x-5)-3(3x-5)]](https://tex.z-dn.net/?f=%28x%2B1%29%5B%202x%283x-5%29-3%283x-5%29%5D)
= 
is the completely factored form of the given polynomial.
Option D is the correct answer.
The range is the difference betweem the largest and smallest NUMBERS in the whole piece of data. The IQR is the difference between the largest and smallest QUARTILES. For example, if you had the data 1,2,3,4,5: 5-1=4 would be the range and...(4-2)=2 would be the IQR. Hope this helped! :)
-8x-12+y^2
This is my aswer haha
