Question

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ᴘʟᴇᴀsᴇ sᴏʟᴠᴇ ᴛʜɪs ǫᴜᴇsᴛɪᴏɴ :

cos²θ - 3cosθ + 2 / sin²θ = 1

(ᴅᴏɴ'ᴛ sᴘᴀᴍ ! ) ​

Answer 1

Step-by-step explanation:

cos²θ – 3cosθ + 2 = sin²θ = 1 – cos²θ

⇒ 2cos²θ – 3cosθ + 1 = 0

⇒ 2cos²θ – 2cosθ – cosθ + 1 = 0

⇒ 2cosθ (cosθ – 1) – 1 (cosθ – 1) = 0

⇒ (2 cosθ – 1) (cosθ – 1) = 0

⇒ cosθ =

1

= 1

2

as cosθ ≠ 1 as θ > 0

∴ cosq = cos 60°

⇒ θ = 60°

Answer 2

Answer:

e = 60°,0°

Step-by-step explanation:

Given Equation is (cos²0 - 3coso + 2)/ sin?0 = 1

→ cos?x 3cosx + 2 = sin?0

- cos?x - 3coso + 2 = (1 - cos?0)

→ cos?0 - 3 cos + 2 -1+ cos20 = 0

→ 2cos?0 - 3cose + 1 = 0

→ 2cos20 - 2 cos - cos + 1= 0

► 2cos(coso - 1) - (coso - 1) = 0

- (2 cos - 1)(coso - 1) = 0

→ cos = (1/2) (or) code = 1 =

- e = 60° (or) 0°

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