Find the number of solution of `sin^2xcos^2x=1+cos^2xsin^4x`
in the interval `[0,2pi]dot`
Text Solution
Verified by Experts
`sin^(2)x cos^(2)x=1+cos^(2)x sin^(4)x` `:. Sin^(2)x cos^(2)x(1-sin^(2) x)=1` `:. Sin^(2) x cos^(4)x=1` `:. Sin^(2)x=cos^(4)x=1`, which is not possible.
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