Q72 of 76 Page 52

True and False

One value of θ which satisfies the equation sin4θ – 2sin2θ – 1 lies between 0 and 2π.

False

Explanation:


We have, sin4θ – 2sin2θ – 1 = 0


Now,




⇒ sin2θ = 1±√2


⇒ sin2θ = 1+√2 or 1-√2


We know that, -1≤ sin θ ≤ 1


⇒ sin2θ ≤ 1


but sin2θ = 1+√2 or 1-√2


Which is not possible.


Hence, the above statement is false.


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