solve for the TWO values of x that satisfy the equation

`(1/16)csc^2(x/4)-cos^2(x/4)=0` on 0° < x < 360°

express exact solutions in degrees.

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`1/16csc^2(x/4)-cos^2(x/4)=0`

Since by definition cosecant is `csc z=1/sin z` we have

`1/(16sin^2(x/4))=cos^2(x/4)`

`16sin^2(x/4)cos^2(x/4)=1`

`(4sin(x/4)cos(x/4))^2=1`

Since `2sin z cos z=sin(2z)` we have

`(2sin(x/2))^2=1`

Now we take quare root and get two solutions

**1)** `2sin(x/2)=1`

**2)** `2sin(x/2)=-1`

**1)** `sin(x/2)=1/2`

`x/2=-30^o +k360^o`

`x=-60^o + k720^o` <-- First solution

`x/2=210^o + k360^o`

`x=420^o +k720^o` <-- Second solution

**2)** `sin(x/2)=-1/2`

`x/2=30^o +k360^o`

`x=60^o + k720^o" <--Third solution"`

`x/2=150^o +k360^o`

`x=300^o +k720^o" <-- Fourth solution"`

Now since first and second solution don't lie between 0° and 360° **you have only two solutions and they are 60° and 300°.**

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