The domain & range of the function will be the answer choice B
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]
Answer:
x=2
Step-by-step explanation:
5×2=10-2=8
You subtract add two to both sides, giving you 5x=10
Divide the 10 by 5 and you get 2.
Answer:
24 sq cm
Step-by-step explanation:
According to theorem of area if parallelogram and triangle, if the triangle and a parallelogram lie between two parallel line and their base is same then, the area of triangle is half of the area of the parallelogram. so here the parallelogram PTRS and the triangle PTR lie between same parallel lines PT and RS and base is also same PT. so area of PTR is 12sq cm. Again, triangles PTR and TRQ lie between same parallel lines PQ and SR and same base PQ. so their area also same that is 12 sq cm. Now, area of PQR = area of PTR + area of TRQ = 12 + 12 sq cm = 24 sq cm.
