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

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Answer 1 · 2016-05-24T10:07:23

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

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

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,

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

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

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