Solve sin^x+sin^2x=1?

1 Answer
Mar 9, 2018

#=>x=kpi+(-1)^k*alpha,kinZ,# where, #alpha=sin^(-1)((sqrt(5)-1)/4)#

Explanation:

#sinx+sin^2x=1=>sin^2x+sinx+1/4=1+1/4=5/4#
#(sinx+1/2)^2=(sqrt(5)/2)^2=>sinx+1/2=+-sqrt(5)/2#
#:.sinx=sqrt(5)/2-1/2 or sinx=-sqrt(5)/2-1/2#
#=>sinx=(sqrt(5)-1)/2 to (1) or sinx=-(sqrt(5)+1)/2 to (2)#
#=>sinx~~(2.24-1)/2=0.62in[-1,1] or##sinx~~-(2.24+1)/2=-1.62!in[-1,1]#
So,
#sinx=sin(sin^(-1)((sqrt(5)-1)/4))#, let, #alpha=sin^(-1)((sqrt(5)-1)/4)#
#=>x=kpi+(-1)^k*alpha,kinZ,# where, #alpha=sin^(-1)((sqrt(5)-1)/4)#