Question 278839
Solve for B to the nearest tenth of a degree in interval odegrees<= B<= 360degrees forth equation 
3(1-sin^2 B) = sin B 
I tried this
The quadratic equation- which left me with 1 +/- square root 37/ 6 if this is correct- how am i supposed to enter this into the calculator- if it is not right then where did I go wrong?

3(1 - sin^2 B) = sin B 
1 - sin^2 B = 1/3 * sin B
sin^2 B - 1 = -1/3 * sin B (divided both sides by -1)
sin^2 B + 1/3 * sin B - 1 = 0
replace sin B with x
x^2 + 1/3 * x - 1 = 0
{{{ x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ x = (-1/3 +- sqrt( 1/9 + 4 ))/2 }}}
{{{ x = -1/6 +- sqrt(37)/6 }}}
{{{ x = (-1 +- sqrt(37))/6 }}}
{{{ x1 = 0.84712709 }}} (approximate)
{{{ x2 = -1.18046042 }}} (approximate)
since x = sin B
we need to solve sin^(-1) x = B
x2 won't work since x needs to be between -1 and 1
solving with x1
B = 57.9 degrees (approximate)