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....
Answer:
168
Step-by-step explanation:
7*-4=-28
-28*-6=168
This is a neat little question. I don't think I've seen it before.
Step one
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Find c
3^2 + 4^2 = c^2
9 + 16 = c^2
c^2 = 25
c = 5
Step 2
====
Set up your first equation for b^2
a^2 + 4^2 = b^2 from triangle XWY
Step 3
=====
Set up your second equation for b^2
25 +b^2 = (a + 3)^2 from triangle XWZ
Step 4
=====
Put the results of Step 2 into step 3 and solve
25 + a^2 + 16 = (a + 3)^2 Collect the like terms on the left.
41 + a^2 = (a + 3)^2 Expand the brackets on the right
41 + a^2 = a^2 + 6a + 9 Transfer the 9 to the left.
32 + a^2 = a^2 + 6a Subtract a^2 from both sides.
6a = 32 Divide by 6
a = 32 / 6
a = 5 2/6
a = 5 1/3