How do you solve #4sin^2 theta - 13sin theta +3 = 0#?

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Answer 1 · 2015-04-30T14:43:54

In this way:

let #t=sinx#,

so:

#4t^2-13t+3=0#

#Delta=b^2-4ac=169-48=121=11^2#

#t_(1,2)=(-b+-sqrtDelta)/(2a)=(13+-11)/8rArr#

#t_1=(13+11)/8=3#

#t_2=(13-11)/8=2/8=1/4#.

And now:

#sinx=3# impossible because the range of the sinus function is #[-1,1]#,

#sinx=1/4rArr#

#x=arcsin(1/4)+2kpi#

#x=pi-arcsin(1/4)+2kpi#.

This is because the solutions of an equation in the variable sinus has (when it has!) 2 supplementary solutions.