How do you solve #3 cos^2 X + 7 sin X = 5#?

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Answer 1 · 2016-05-22T14:18:57

General solution of #3sin^2X-7sinX+2=0# is

#npi+(-1)^npi/9.244#

#3cos^2X+7sinX=5#

or #3(1-sin^2X)+7sinX-5=0#

or #-3sin^2X+3+7sinX-5=0#

or #3sin^2X-7sinX+2=0#

or #3sin^2X-6sinX-sinX+2=0#

or #3sinX(sinX-2)-1(sinX-2)=0#

or #(3sinX-1)(sinX-2)=0#

But as range of sinX is within #[-1,1]#

#sinX-2!=0# and hence

#3sinX-1=0# or #sinX=1/3# or #X=19.47^o=pi/9.244#

Hence General solution of #3sin^2X-7sinX+2=0# is

#npi+(-1)^npi/9.244#