Trigonometrical Equations 1 Question 24

24. There exists a value of θ between 0 and 2π that satisfies the equation sin4⁡θ−2sin2⁡θ+1=0.

(1984, 1M)

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Answer:

Correct Answer: 24. False

Solution:

  1. Given, sin4⁡θ−2sin2⁡θ+1=2

⇒(sin2⁡θ−1)2=2⇒sin2⁡θ=±2+1

which is not possible. Hence, given statement is false.

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