Question

Plz solve for θ :-

2sin²θ= 3cosθ ; where 0≤θ≤2π

{Class XI Trigonometry}​

Answer 1

Step by step explanation.
2(1-cos²A) = 3CosA
2-2cos²A = 3 Cos A
2= 3 cosA + Cos²A
2= cos A ( 3+ cos A)
now,
either cos A = 2
or 3 + cos A = 2
cos A = 2-3
or Cos A = -1
so. A = 180

Answer 2

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