There exists a value of `theta` between 0 and `2pi` that satisfies the equation `sin^4theta-2sin^2theta-1=0`

Explanation

`sin^4theta-2sin^2theta-1=0`

Let `sin^2theta=y`

`y^2-2y-1=0`

`y=(-(-2)pmsqrt(4-4(1)(-1)))/2`

`y=(2pmsqrt8)/2`

`y=(2pm2sqrt2)/2`

`sin^2theta=1pmsqrt2`

`sin^2theta=1+sqrt2`

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