La longitud del arco (s) en una circunferencia, conociendo el radio (r) y el ángulo (θ) que forman los dos radios, es:
s = r∙θ
Con el ángulo en radianes
F V7 w7 :
If you look at your graph you see that there is a hole at x = 3. This results from a removeable discontinuity. That means that when we factor the numerator to (x-3)(x-1), that (x-3) cancels out between the numerator and denominator and a hole is what results from that cancelation. The other factor for the denominator is found where the vertical asymptote of the function is, which is found at x = -4. That means that the 2 factors for the denomonator are (x-3) and (x+4). When you FOIL those out, you get . That means that a = 1 and b = -12
Answer with explanation:
The equation using trigonometric function are whose value we have to evaluate in the interval [0, 2π].
1.→
cos²x +2 cos x=3
→cos²x +2 cos x-3=0
→cos²x +3 cos x-cos x-3=0
→cos x ( cos x +3) -1×(cos x +3)=0
→ (cos x -1)(cos x +3)=0
→cos x -1= 0 ∧ →cos x+3=0
→cos x =1 ∧ → cos x = -3
⇒-1 ≤ cos x ≤ 1
→cos x=1
where, n is any integer.
→Solution in the Interval [0, 2π] is , x=0, 2 π
2.
2 sin² Ф +7 sin Ф =4
→2 sin² Ф +7 sin Ф -4=0
→2 sin² Ф +8 sin Ф- sin Ф -4=0
→2 sin Ф × (sin Ф +4) -1 × (sin Ф +4)=0
→ (2 sin Ф -1)(sin Ф +4)=0
→ 2 sin Ф -1=0 ∧ →sin Ф +4=0
→sin Ф ≠ -4≡→ ∵ -1 ≤ sin Ф ≤ 1
Where , n is any Integer.
3.
→tan² x+4=2 sec² x+tan x
→tan² x+4=2(1+tan²x)+tan x
→2 tan²x -tan²x +tan x +2-4=0
→tan²x +tan x -2=0
→tan²x +2 tan x -tan x -2=0
→tan x (tan x +2) -1×(tan x +2)=0
→(tan x-1)(tan x +2)=0
→tan x-1 =0 ∧ tan x +2=0
→tan x=1 ∧ tan x = -2
→tan x=1
→tan x= -2
where, k and n are integer.