Question

How do you solve `sin(1/2x)+cos(x)=1` in the interval [0, 2pi]?

Answer 1

`x=0, pi/6, (5pi)/6 and 2pi in [0, 2pi]`..

Explanation:

`sin (x/2)+cos x= sin (x/2)+1-2 cos^2(x/2)=1`.

So, `sin (x/2)(1-2 sin (x/2))=0`

The factor `sin (x/2)=0`, gives `x=2npi, n=0,+-1,+-2,+-3,.`.

and the other factor 1-2sin (x/2)=0`gives`sin (x/2)=1/2#, and so,

`x=2(npi+(-1)^npi/6), n= 0, +-1, +-2, +-3,..`

`x=0, pi/6, (5pi)/6 and 2pi,in [0, 2pi]`.,