Solve the following inequality:
(i) `sin x ge sqrt(3)/2`
(ii) `sinx + cosx ge 1`

Explanation

To solve the given inequality we draw the graph of y= `sinx` and `y=sqrt(3)/2`, clearly if `pi/3 le x le (2pi)/3` in`[0,2pi]`, then `sinx ge sqrt(3)/2`. As the period of `sinx` is `2pi`,hence general solution is `2npi + pi/3 le x le 2npi + (2pi)/3, n int Z`
(ii) `sinx + cosx ge 1`
`rArr 1/sqrt(2) sinx + 1/sqrt(2) cosx ge 1/sqrt(2)`
`rArr sin(x+pi/4) ge 1/sqrt(2)`
`rArr pi/4 le x +pi/4 le (3pi)/4`
`rArr 0 le x le pi/2`
As the period of `sinx + cosx` is `2pi`, hence general solution is `[2npi, 2npi +pi/2], n int Z`.