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....
Step-by-step explanation:
Hi!
I can help you out!
16/20=4/5=8/10=32/40
Hope this can help, choose whichever fraction, they're all the same {it just depends the options you're given
Hope this helped! <3
Answer:
73
Step-by-step explanation:
2×2×7×2=56
2+2=4
56+4+13=73
it's a parallel line so the gradient stays the same
y= x + c
substitute in the point specified
2 = 2 + c
2-2=0
so the answer is y=x
hope this helps!
C. is the answer cuz if u do a number line and organize it and put the numbers in there places the right side is the biggest the ones in the left towards the negative is the lowest
HOPE THIS HELPS:)
