How can this be solved?

Question

#cos(x/2) - sinx = 0#

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Answer 1 · 2018-02-28T05:14:06

#"The Solution Set="{(4k+-1)pi}uu{2kpi+(-1)^k*pi/3}, k in ZZ#.

Prerequisites :#(1):costheta=cosalphahArrtheta=2kpi+-alpha, k in ZZ#.

#(2):sintheta=sinalphahArrtheta=kpi+(-1)^kalpha, k in ZZ#.

#cos(x/2)-sinx=0#.

#:. cos(x/2)-2sin(x/2)cos(x/2)=0#.

#:. cos(x/2){1-2sin(x/2)}=0#.

#:. cos(x/2)=0, or, sin(x/2)=1/2#.

Now, #cos(x/2)=0=cos(pi/2)#,

#:. x/2=2kpi+-pi/2 rArr x=4kpi+-pi, k in ZZ............(i), and, #

#sin(x/2)=1/2=sin(pi/6)rArr x/2=kpi+(-1)^k*pi/6#,

#rArr x=2kpi+(-1)^k*pi/3, k in ZZ............(ii)#.

Combining #(i) and (ii)#, we get the desired

#"The Solution Set="{(4k+-1)pi}uu{2kpi+(-1)^k*pi/3}, k in ZZ#.