Math, asked by unknown01011947, 6 months ago

2√3cos²θ = sinθ
find the value of θ

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Answers

Answered by adishpandya
0

Answer:

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Answered by hykokcha
1

Answer:

2√3 cos² θ = sin θ

⇒ cos² θ / sinθ = 1/2√3                            

⇒ (1 - sin²θ) / sinθ = 1/2√3

                                                                     

Squaring both sides

⇒ (1 - sin²θ)² / sin²θ = (1/2√3)²

⇒ (1 - sin²θ)² / sin²θ = 1/12

⇒ 12 ( 1² + [sin²θ]² - 2sin²θ ) = sin²θ

⇒ 12 ( 1 + sin^{4}θ - 2sin²θ ) = sin²θ

⇒ 12 + 12sin^{4}θ - 24sin²θ = sin²θ

⇒ 12 + 12sin^{4}θ - 24sin²θ - sin²θ = 0

⇒ 12sin^{4}θ - 25sin²θ + 12 = 0

                                                      Taking sin²θ = y

⇒ 12y² - 25y + 12 = 0

⇒ 12y² - 16y - 9y + 12 = 0

⇒ 4y(3y-4) - 3(3y-4) = 0

⇒ (4y-3) (3y-4) = 0

Therefore, y = 3/4 Or

                  y = 4/3

So, sin²θ = 3/4 Or

     sin²θ = 4/3

                                                  Taking sin²θ = 4/3

⇒     sin²θ = 4/3

⇒     sinθ = √4/√3

⇒     sinθ = 2/√3           Ignoring this as this will

not give the value of θ              

Therefore, taking sin²θ = 3/4

⇒ sin²θ = 3/4

⇒ sinθ = √3/√4

⇒ sinθ = √3/2

                 

And we know that sin60° = √3/2

Therefore, the value of θ = 60°

Hope it helps you .....

XD

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