How do you use the quadratic formula to solve #2cot^2theta+3cottheta-4=0# for #0<=theta<360#?

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Answer 1 · 2017-07-19T09:46:39

The solutions are #S={49.61^@,156.57^@,229.61^@}#

We solve this like the quadratic equation

#ax^2+bx^2+c=0#

Our equation is

#2cot^2x+3cotx-4=0#

The discriminant is

#Delta=b^2-4ac=3^2-4*(2)*(-4)=9+32=41#

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

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

#cotx=(-3+-sqrt41)/(4)#

The solutions are

#x_1=arc cot((-3+sqrt41)/(4))=arc cot(0.8508)=arctan(1/0.8508)=arctan1.1754=49.61^@, 229.61^@#

#x_2=arc cot((-3-sqrt41)/(4))=arc cot(-2.308)=arctan(-1/2.308)=arctan(-0.4333)=-23.43^@=156.57^@#