Find-all-angles-between-0-and-360-for-which-8sin-3cos-2-

Question Number 85489 by oustmuchiya@gmail.com last updated on 22/Mar/20

F i n d a l l a n g l e s b e t w e e n 0 ° a n d 360 °, f o r w h i c h 8 s i n θ = 3 {c o s}^{2} θ

Commented by Tony Lin last updated on 22/Mar/20

8 s i n θ = 3 (1 - {s i n}^{2} θ)

3 {s i n}^{2} θ + 8 s i n θ - 3 = 0

(3 s i n θ - 1) (s i n θ + 3) = 0

s i n θ = \frac{1}{3}

θ ≒ 19.47 ° & 160.53 °

Commented by john santu last updated on 22/Mar/20

8 \sin θ = 3 (1 - \sin^{2} θ)

3 \sin^{2} θ + 8 \sin θ - 3 = 0

(3 \sin θ - 1) (\sin θ + 3) = 0

\sin θ = \frac{1}{3} \Rightarrow θ = {\begin{cases} \sin^{- 1} (\frac{1}{3}) + 2 k π \\ π - \sin^{- 1} (\frac{1}{3}) + 2 k π \end{cases}