How do you find the solution to #3sin^2theta+7sintheta+2=0# if #0<=theta<2pi#?

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Answer 1 · 2017-02-08T16:13:07

The solutions are #S={199.5º, 340.5º}#

We compare this equation to the quadratic equation

#ax^2+bx+c=0#

#3sin^2theta+7sintheta+2=0#

We calculate the discriminant

#Delta=b^2-4ac=(7^2)-4(3)(2)=49-24=25#

As, #Delta>0#, there are 2 real solutions

#sintheta=(-b+-sqrtDelta)/(2a)#

#=(-7+-sqrt25)/(2*3)#

#=(-7+-5)/(6)#

#sintheta_1=(-7+5)/6=-2/6=-1/3#

#sintheta_2=(-7-5)/6=-12/6=-2#

This is impossible since #-1<=sintheta<=1#

So,
#theta_1=arcsin(-1/3)#

#theta_1=199.5º# and #340.5º#