Find the number of solutions of the equation `sin^4x+cos^4x-2sin^2x+3/4sin^2 2x=0` in the interval `[0,2pi]`

Answer

Four solutions

Explanation

`sin^(4)x+cos^(4)x-2 sin^(2)x+3/4 sin^(2) 2 x=0`
`rArr (sin^(2)+cos^(2) x)^(2)-2 sin^(2)x cos^(2)x-2 sin^(2) x +3/4 . 4 sin^(2) x. cos^(2) x=0`
`rArr 1-2 sin^(2) x+sin^(2) x cos^(2) x=0`
`rArr sin^(4)x + sin^(2) x-1=0`
`rArr sin^(2) x=(sqrt(5)-1)/2`
`rArr sin x= pm sqrt((sqrt(5)-1)/2)`
Thereare two values of x for `sin x= sqrt((sqrt(5)-1)/2)` and two values of `x` for `sin x=-sqrt((sqrt(5)-1)/2)`.
so, there are four solutions.