Question

(a) Solve 3cosθ+2=0 for 0°≤θ≤360°.
--->132°, 228°
(b) Hence solve 3cos^(2)θ+2cosθ=0 for 0°≤θ≤180°.

@hartnn

Answer 1

i don't know how to do part b.

Answer 2

3c^2+2c = 0
c (3c+2) = 0
c=0 or 3c +2=0

Answer 3

so what angle makes cos theta = 0 ?

Answer 4

90, 270(rej)

Answer 5

∴90

Answer 6

one will be rejected from 132°, 228°
too

Answer 7

228

Answer 8

so u'll have 2 angles

Answer 9

oh i see. haha i am so stupid...