(X+2)(y-5)=0
X+2=0
X=-2
Y-5=0
Y=5
X is horizontal
Y is vertical
I) HCF - use the smallest powers of each common factors
HCF (A,B) = 2^2 × 3^4 × 5^2
LCM - use the highest powers of each factors
LCM (A,B) = 2^4 × 3^6 × 5^2 × 7^2 × 11^16
ii) Add powers together.
A×B = 2^6 × 3^10 × 5^4 × 7^2 × 11^16
sqrt(A × B)
Divide powers by 2.
sqrt(A × B) = 2^3 × 3^5 × 5^2 × 7 × 11^8
iii) C = 3^7 × 5^2 × 7
Ck = (3^7 × 5^2 × 7) × k
B/c Ck should be a product that is a perfect cube, the powers of the products should be divisible by 3.
(3^7 × 5^2 × 7) × k = 3^9 × 5^3 × 7^3
k = (3^9 × 5^3 × 7^3) / (3^7 × 5^2 × 7)
k = 3^(9-7) × 5^(3-2) × 7^(3-1)
k = 3^2 × 5 × 7^2
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
