For 0 less than or equal to theta less than 2(pi), what are thebsolutions to sin↑2(theta)=2(sin↑2)(theta/2)? ​

I assume the up arrows are supposed to indicate exponents, so that the equation is

sin²(θ) = 2 sin²(θ/2)

Recall the half-angle identity for sine,

sin²(θ/2) = (1 - cos(θ))/2,

as well as the Pythagorean identity,

sin²(θ) + cos²(θ) = 1

Rewrite the equation in terms of cosine and solve:

1 - cos²(θ) = 1 - cos(θ)

cos²(θ) - cos(θ) = 0

cos(θ) (cos(θ) - 1) = 0

cos(θ) = 0   or   cos(θ) - 1 = 0

cos(θ) = 0   or   cos(θ) = 1

[θ = arccos(0) + 2nπ   or   θ = arccos(0) - π + 2nπ]   or

… … … [θ = arccos(1) + 2nπ]

(where n is any integer)

[θ = π/2 + 2nπ   or   θ = -π/2 + 2nπ]   or   [θ = 2nπ]

In the interval 0 ≤ θ < 2π, we get the solutions θ = 0, π/2, and 3π/2.

(That is, for n = 0 in the first and third solution families, and n = 1 in the second family.)

1.4

2.3

step-by-step explanation:

the formula for the constant of variation is

y= kx

divide each side by x

y/x =kx/x

y/x =k

we know a and y

1.   y/x = k

4/1 =4

check the other values to make sure they are the same

8/2 = 4   12/3 =4   and 16/4 =4  

the constant of variation is 4

2.   y/x = k

3/1 =3

check the other values to make sure they are the same

6/2 = 3   9/3 =3   and 12/4 =3  

the constant of variation is 3

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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