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

How do you solve 3sin^2x=3sinx+23sin^2x=3sinx+2 in the interval [0,360]?