How do you solve #3sin^2x=3sinx+2# in the interval [0,360]?

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Answer 1 · 2017-01-26T14:30:46

The solutions are #S={332.8º, 207.2 º}#

Let's rewrite the equation

#3sin^2x-3sinx-2=0#

We solve this like a quadratic equation

#ax^2+bx+c=0#

We start by calculating the discriminant

#Delta=b^2-4ac=(-3)^2-4*3*(-2)=9+24=33#

#Delta >0#, we have 2 real solutions

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

#sinx=(3+-sqrt33)/(6)#

#sinx=1.46#, there is no solution as #-1<=sinx<=1#

#sinx=-0.457#, #x=332.8#º and #207.2#º