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.
Since by definition cosecant is `csc z=1/sin z` we have
Since `2sin z cos z=sin(2z)` we have
Now we take quare root and get two solutions
`x=-60^o + k720^o` <-- First solution
`x/2=210^o + k360^o`
`x=420^o +k720^o` <-- Second solution
`x=60^o + k720^o" <--Third solution"`
`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°.