Question

Value for θ, for sin2θ=1, where 0∘<θ<90∘ is

Answer 1

Answer:

MATHS

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Asked on December 30, 2019 by

Shafaque Manju

Solve : sin2θ=cos3θ,0

∘

≤θ≤360

∘

.

Which of the following values are a part of the solution (jn degrees)

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ANSWER

sin2θ=cos3θ

⇒2sinθcosθ=4cos

3

θ−3cosθ

⇒2sinθcosθ=4cosθ(1−sin

2

θ)−3cosθ

⇒2sinθcosθ=−4cosθsin

2

θ+cosθ

⇒4cosθsin

2

θ−cosθ+2sinθcosθ=0

⇒cosθ(4sin

2

θ−1+2sinθ)=0

⇒cosθ=0 or 4sin

2

θ+1−2sinθ=0

⇒θ=

2

(2n+1)π

or sinθ=−1±

5

Hence for 0

∘

≤θ≤360

∘

θ=18

∘

,90

∘

,162

∘

,234

∘

,270

∘

,306

∘

Answer By

Shreya pathak

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