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:
<u>Δ MNO ≅ Δ PQR by ASA postulate</u>
Step-by-step explanation:
Δ MNO and Δ PQR are congruents because:
1. Their included sides MN and PR are equal (22 units = 22 units)
2. Their angles ∠M and ∠P are equal (100° = 100°)
3. Their angles ∠N and ∠Q are equal (35° = 35°)
<u>Now, we can conclude that Δ MNO ≅ Δ PQR by ASA postulate.</u>
3wx^2
The third choice is the answer.
Answer :C
B. The answer would be B because inscribed means inside the circle. All the points in the angle touch the side of the circle.
Hope I could help!!!!!
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