# Question 25: Trigonometric equations. (Late Middle Palaeolithic).

QUESTION 25.

TRIGONOMETRIC EQUATIONS (Late Middle Palaeolithic).

θ = s/60 × 2π

8sin^2(θ) + 2sin(θ) – 1 = 0

2cos^2(θ) + √3cos(θ) = 0

a. Solve for θ and s. (Note: there are several possible θ values, find the specific angle that corresponds to both sine and cosine).

8sin^2(θ) + 4sin(θ) – 2sin(θ) – 1 = 0

4sin(θ)(2sin(θ) + 1) – 1(2sin(θ) + 1) = 0

(2sin(θ) + 1) (4sin(θ) – 1) = 0

2sin(θ) + 1 = 0

2sin(θ) = -1

sin(θ) = -1/2

4sin(θ) – 1 = 0

4sin(θ) = 1

sin(θ) = 1/4

cos(θ)(2cos(θ) + √3) = 0

cos(θ) = 0 or 2cos(θ) + √3 = 0

cos(θ) = 0 or cos(θ) = -√3/2

sin^-1(-1/2) = π – θ = π/6

θ = π – sin^-1(-1/2) = 7π/6

cos^-1(-√3/2) = 2π – θ = 5π/6

θ = 2π – cos^-1(-√3/2) = 7π/6

7π/6 = s/60 × 2π

7/12 = s/60

s = 7/12 × 60 = 35

b. Re-divide s by 60 to get s/60. Then choose k for minutes (m) and add k + s/60 to get m. Then divide m by 60 and choose k for hours (h) then add k + m/60 to get h. Then divide h by 24 and choose k for days (d) then add k + h/24 to get d. Then divide d by 30 and choose k for months (M) then add k + d/30 to get M. Then divide M by (73/6) and choose k for years (y) then add k + M/(73/6) to get y.

Tip: choose a meaningful k for years (y), especially if you have sinned, such as the Late Middle Palaeolithic, 60,000 to 35,000 years ago. For example, if you are a cannibal, this is a place and time where you can find relative innocence. For a full understanding of this read https://PhDForgive.com.

s/60 = 35/60 = 7/12

k = m – s/60 = 39

m = k + s/60 = 475/12

m/60 = 95/144

k = h – m/60 = 19

h = k + m/60 = 2831/144

h/24 = 2831/3456

k = d – h/24 = 23

d = k + h/24 = 82319/3456

d/30 = 82319/103680

k = M – d/30 = 7

M = k + d/30 = 808079/103680

M/(73/6) = 808079/1261440

k = y – M/(73/6) = 50350

y = k + M/(73/6) = 63514312079/1261440

c. Choose and workout the magnitudes of T^1, T^-1, T^2 and T^-2, also workout t, A and B.

T^1 = √(A/B) = 10^18 as

T^-1 = √(B/A) = 10^-18 Es

T^2 = A/B = 10^36 as²

T^-2 = B/A = 10^-36 Es²

t = A/T^1 = y × 31536000 = 1.587857801975 × 10^12 s

A = tT^1 = X × 10^12 × 10^18 = X × 10^30 as

B = A/T^2 = X × 10^30 / 10^36 = X × 10^-6 Es

d. Although you have done it forward or ascending, undo or reverse-workout the integers and decimals of the different time units backward or descending.

M/(73/6) = y – k = 808079/1261440

M = (y – k) × (73/6) = 808079/103680

k = M – d/30 = 7

d/30 = M – k = 82319/103680

d = (M – k) × 30 = 82319/3456

k = d – h/24 = 23

h/24 = d – k = 2831/3456

h = (d – k) × 24 = 2831/144

k = h – m/60 = 19

m/60 = h – k = 95/144

m = (h – k) × 60 = 475/12

k = m – s/60 = 39

s/60 = m – k = 7/12

s = (m – k) × 60 = 35

e. Check months with days.

y – d/365 = t / 31536000 – d/365 = 50350

d – h/24 = (y – k) × 365 – h/24 = 233

h – m/60 = (d – k) × 24 – m/60 = 19

m – s/60 = (h – k) × 60 – s/60 = 39

s – cs/100 = (m – k) × 60 – cs/100 = 35

50350 years 233 days 19:39:35

50350 years 7 months 23 days 19:39:35

Hand written example:

The Late Middle Paleolithic, about 60,000 to 35,000 years ago. Again, for example, if you are a cannibal, this is a place and time where you can find relative innocence. For a full understanding of this read https://PhDForgive.com.