1.

Find the least positive value of `x`satisfying`(sin^2 2x+4sin^4x-4sin^2xcos^2x)/4=1/9`

Answer» `(sin^2 2x+4sin^4x - 4sin^2xcos^2x)/4 = 1/9`
`=>((2sinxcosx)^2+4sin^4x - 4sin^2xcos^2x)/4 = 1/9`
`=>(4sin^2xcos^2x+4sin^4x - 4sin^2xcos^2x)/4 = 1/9`
`=>(4sin^4x)/4 = 1/9`
`=>sin^4x = 1/9`
`=>sinx = +-1/sqrt3`
So, least positive value of `x` will be `sin^-1(1/sqrt3).`


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