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Question Number 47586 by Rio Michael last updated on 11/Nov/18

G i v e n t h e f u n c t i o n f (θ) = c o s 2 θ - s i n θ . (f o r 0 ° ⩽ θ ⩽ π)

p l o t t h e g r a p h f o r i n t e r v a l s o f \frac{π}{6} .

h e n c e f i n d t h e v a l u e o f c o s 2 θ = s i n θ .

Answered by Joel578 last updated on 12/Nov/18

Commented by Joel578 last updated on 12/Nov/18

\cos 2 θ = \sin θ

1 - {2 \sin}^{2} θ = \sin θ

{2 \sin}^{2} θ + \sin θ - 1 = 0

(2 \sin θ - 1) (\sin θ + 1) = 0

\sin θ = \frac{1}{2} \lor \sin θ = - 1

Commented by Rio Michael last updated on 13/Nov/18

t h a n k s s i r . B u t i t h o u g h t

c o s 2 θ = s i n θ

c o u l d b e l o c a t e d f r o m t h e g r a p h . T h a t i s a t p o i n t w h e r e

c o s 2 θ - s i n θ (f (θ)) = 0