Question

Solve cos^2 (B) + 3cos(B) + 2 = 0 for B, 0 < B < 360 Please help

Answer 1

Step-by-step explanation:

$\cos {}^{2} (b) + 3 \cos(b) + 2 = 0$

$( \cos(b) + 2)( \cos(b) + 1) = 0$

$\cos(b) = - 2$

Cosine have a range of [-1,1] so b is undefined for the first factor

$\cos(b) = - 1$

$b = 180$