Solve the equation for all degree solutions and if 0° ≤ θ < 360°. do not use a calculator (enter your answers as a comma-separated list. if there is no solution, write no solution.)

1. 2 sin θ= root 2
(a) all degree solutions (let k be any integer.)
(b) 0≤ θ < 360°

2. 2 cos θ − root 3=0

(a) all degree solutions (let k be any integer.)
(b) 0≤ θ < 360°

3. root 3 cot θ- 1=0

(a) all degree solutions (let k be any integer.)
(b) 0° ≤ θ < 360°

1.-    

a)      θ =  45 ° + 360k

         θ =  135° + 360k

b)        θ =  45 °             0 ≤  θ < 360°

          θ =  135°  

2.-    

a)          θ  =  30° + 360k

            θ  = 330° + 360k

b)         θ  =  30°      0  ≤  θ  < 360°

           θ  = 330°

3.-        

a)         θ    =  60°  + 360k

            θ   =   240° + 360k

b)         θ    =  60°           0  ≤   θ < 360°

            θ    = 240°  

Step-by-step explanation:

1)    2 sin θ  = √2

   sin θ   =  √2/2

1.a )   θ =  45 ° + 360k

         θ =  135° + 360k

1.b)     θ =  45 °             0 ≤  θ < 360°

          θ =  135°  

2.

2  cos  θ  -  √3   =  0

2  cos  θ  = √3  

    cos  θ  = √3/2

2.a        θ  =  30° + 360k

            θ  = 330° + 360k

2.b     θ  =  30°      0  ≤  θ  < 360°

           θ  = 330°

3.-     √3 cot  θ  -  1  = 0

         √3 cot  θ    =  1

              cot  θ    =  1 /√3

3.a          θ    =  60°  + 360k

               θ   =   240° + 360k

3.b           θ    =  60°           0  ≤   θ < 360°

                θ    = 240°  

step-by-step explanation:

c) the median for town a 30" is greater than the median for town b 25"

step-by-step explanation:

town a, median = q2 = 30

town b, median - q2 = 25

so the median for town a 30" is greater than the median for town b 25"