How do you use the quadratic formula to solve #sintheta/2=3/(sintheta+2)# for #0<=theta<360#?

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Answer 1 · 2017-03-29T06:46:53

There is no solution.

Note that if #a/b=c/d# then #axxd=bxxc#

Now we have #sintheta/2=3/(sintheta+2)#, hence

#sintheta(sintheta+2)=6#

or #sin^2theta+2sintheta-6=0#

Now as per quadratic formula, if we have a quadratic equation #ax^2+bx+c=0# in #x#, then #x=(-b+-sqrt(b^2-4ac))/(2a)#

Here we have the equation #sin^2theta+2sintheta-6=0# in #sintheta# and therefore using quadratic formula

#sintheta=(-2+-sqrt(2^2-4*1*(-6)))/2# or

#=(-2+-sqrt(28))/2=-1+-sqrt7#

But #sintheta# can have values only from #-1# to #1# and both values #-1+sqrt7# and #-1-sqrt7#, are out of the range. Hence

There is no solution.

Answer 2 · 2017-03-29T06:48:22

no solution. see explanation.

#sin theta/2 = 3/(sin theta +2)#

#sin theta(sin theta +2) = 3 *2#
#sin^2 theta + 2sin theta = 6#
#sin^2 theta + 2sin theta - 6 = 0#

#a = 1, b = 2 and c= -6#

#sin theta = (- b +-sqrt(b^2 - 4ac))/(2a)#

#sin theta = (- 2 +-sqrt(2^2 - 4(1)(-6)))/(2(1))#

#sin theta = (- 2 +-sqrt(4 +24))/(2)#

#sin theta = (- 2 +-sqrt(28))/(2) = (- 2 +-sqrt(4*7))/(2)#

#sin theta = (- 2 +-2sqrt(7))/(2)= -1 +-sqrt(7) = -1 +- 2.656#

when #sin theta = -1 + 2.656 = 1.656,#
#theta = sin^-1 (1.656)# no solution/not valid

when #sin theta = -1 - 2.656 = -3.656,#
#theta = sin^-1 (-3.656)# no solution/not valid