For 0 ≤ θ < 2 π what are the solutions to sin^2(θ) =2sin^2(θ/2)

Recall the half-angle identity for sine,

sin²(x/2) = (1 - cos(x))/2

Then the given equation is identical to

sin²(θ) = 1 - cos(θ)

Also recall the Pythagorean identity,

sin²(θ) + cos²(θ) = 1

Then we rewrite the equation as

1 - cos²(θ) = 1 - cos(θ)

Factoring the left side, we have

(1 - cos(θ)) (1 + cos(θ)) = 1 - cos(θ)

and so

(1 - cos(θ)) (1 + cos(θ)) - (1 - cos(θ)) = 0

and we factor this further as

(1 - cos(θ)) (1 + cos(θ) - 1) = 0

which gives

cos(θ) (1 - cos(θ)) = 0

Then either

cos(θ) = 0   or   1 - cos(θ) = 0

cos(θ) = 0   or   cos(θ) = 1

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

…   or   [θ = arccos(1) + 2nπ   or   θ = -arccos(1) + 2nπ]

(where n is any integer)

[θ = π/2 + 2nπ   or   θ = -π/2 + 2nπ]   or   [θ = 0 + 2nπ]

In the interval 0 ≤ θ < 2π, we get three solutions:

• first solution set with n = 0   ⇒   θ = π/2

• second solution set with n = 1   ⇒   θ = 3π/2

• third solution set with n = 0   ⇒   θ = 0

So, the first choice is correct.

$655.61 which is option c

step-by-step explanation:

price of house =   $146,000

initial rate = 3.5%

mortgaged paid in this = 3.5 % of $146,000

                                        =

                                        =$5110

the total years would be = 7

total mortgaged paid = 7 * $5110

                                      =$35770

remaining money = $146000-35770

                                =$110230

new mortgaged payment = 5 % of 110230

                                            =  

                                            =$655.61

here's a poem i finished a while ago. i hope you like it and that it inspires you.