How do you use the quadratic formula to solve #9sin^2theta+6sintheta=2# for #0<=theta<360#?

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Answer 1 · 2017-08-08T06:32:55

The solutions are #S={14.1^@, 165.9^@, 245.6^@, 294.4^@}#

The equation is

#9sin^2theta+6sintheta=2#

#9sin^2theta+6sintheta-2=0#

Comparing this equation to the quadratic equation

#ax^2+bx+c=0#

We see that #x =sintheta#

We start by calculating the discriminant

#Delta=b^2-4ac=6^2-4(9)(-2)=36+72=108#

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

#sintheta=(-b+-sqrtDelta)/(2a)=(-6+-sqrt108)/(18)=(-6+-6sqrt3)/(18)#

#sintheta=-1/3+sqrt3/3#, #=>#, #theta=14.1^@# or #165.9^@#

#sintheta=-1/3-sqrt3/3#, #=>#, #theta=294.4^@# or #245.6^@#