How do you find the exact value of #3-3sintheta-2cos^2theta=0# in the interval #0<=theta<360#?

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Answer 1 · 2017-04-25T13:49:20

The solutions are #S={30º,90º,150º}#

We need

#sin^2theta+cos^2theta=1#

#cos^2theta=1-sin^2theta#

We simplify the equation

#3-3sintheta-2cos^2theta=0#

#3-3sintheta-2(1-sin^2theta)=0)#

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

#2sin^2theta-3sintheta+1=0#

We compare this equation to the quadratic equation

#ax^2+bx+c=0#

The discriminant is

#Delta=b^2-4ac=(-3)^2-4(2)(1)=9-8=1#

As,

#Delta>0#, there are 2 real solutions

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

#=(3+-1)/4#

Therefore,

#sintheta=4/4=1#, #=>#, #theta=90º#

#sintheta=2/4=1/2#, #=>#, #theta=30ª, 150ª#