SOLUTION: 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 wi

Algebra ->  Trigonometry-basics -> SOLUTION: 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 wi      Log On


   

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?
Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
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+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+x+=+%28-1%2F3+%2B-+sqrt%28+1%2F9+%2B+4+%29%29%2F2+
+x+=+-1%2F6+%2B-+sqrt%2837%29%2F6+
+x+=+%28-1+%2B-+sqrt%2837%29%29%2F6+
+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)