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Question

Write the values of x in [0, π] for which sin 2x,12 and cos 2x are in A.P.

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Solution

(i)
sin2x, 12 and cos2x are in AP.sin2x+cos2x=2×12sin2x+cos2x=1 ...(1)

This equation is of the form a sinθ + b cosθ = c, where a = 1, b = 1 and c = 1.
Now,
Let:
a = r sin α and b = r cos α
Thus, we have:
r = a2 + b2 = 12+12 = 2 and tan α = 1 α = π4
On putting a = 1 = r sin α and b = 1 = r cos α in equation (1), we get:
r sin α sin2x + r cosα cos2x = 1
r cos (2x - α) = 1 2 cos 2x - π4 = 1 cos 2x- π4 = 12 cos 2x-π4 = cos π4 2x - π4 = 2nπ ± π4 , n ZTaking positive value, we get: 2x - π4 = 2nπ + π4x = nπ+π4Taking negative value, we get: 2x - π4 = 2nπ - π42x - π4 = 2nπ - π4 x =nπ, n Z

For n = 0, the values of x are π4 and 0 and for n = 1, the values of x are 5π4 and π.
5π4 does not satisfy the condition.
For the other value of n, the given condition is not true, i.e., [0, π].

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