How do you solve for x given #cos 58=sqrt((1+cosx)/2)#?

1 Answer

#x = (720k+- 116)^o, k = 0, +-1, +-2, +-3...#.
In #(0, 360^o)#, ( for #k=0# and + sign) #x = 116^0 and 244^o# (equivalent to #-116^o#. for #k = 0# and negative sign )...

Explanation:

#sqrt((1+cosx)/2)=sqrt(cos^2(x/2)#

#=cos(x/2)=cos 58^o#

The general solution is

#x/2 = (360k+- 58)^o, k = 0, +-1, +-2, +-3...#. So,

#x = (720k+- 116)^o, k = 0, +-1, +-2, +-3...#.

In #(0, 360^o)#, ( for #k=0# and + sign) #x = 116^o and 244^o# (equivalent to #-116^o#, for #k = 0# and negative sign )...