Find the range of sinx satisfying ` log _(" cosx") sinx ge 2 ` .

Explanation

We have ,
` log_(" cosx")sinx ge 2 `
The domain of admissble values is
` sin x gt 0 and cosx gt 0 , cosx ne 1`
i.e. ` 0 lt sinx le 1 and - lt cosx lt 1`
The equivalent form of the given inequality is ` sin x le (cos x )^(2) = 1 - sin^(2) x `
`rArr sin^(2) x + sinx - 1 le 0 `
`rArr ( sin x + (1)/(2))^(2) le (1)/(4) + 1 (5)/(4)`
` rArr sin x + (1)/(2) le (sqrt(5))/(2) or ge - (sqrt(5))/(2)`
But ` sin x gt 0 `
` therefore sin x le (sqrt(5)-1)/(2)`
Hence , the range of sin ` x is (0, (sqrt(5) -1)/(2))` .