Trigonometric EquationHard
Question
If cot(3α - 7β) tan (5β - α) = 1, α∈(0, π), β∈
and L = 2√2 sin 
, then -
and L = 2√2 sin , then - Options
A.Number of integral values of L is 2
B.Number of integral values of L is 3
C.Number of irrational values of L is 4
D.Number of irrational values of L is 3
Solution
cot(3α - 7β) tan(5β - α) = 1
⇒ cos(3α - 7β) sin (5β - α)
= sin(3α - 7β) cos(5β - α)
⇒ sin[(5β - α) - (3α - 7β)] = 0
⇒ sin(4α - 12β) - (3α - 7β)] = 0
⇒ sin(4α - 12β) = 0
⇒ 4(α - 3β) = nπ ⇒
∵ α - 3β ∈ (-π, π)
⇒
⇒ - 4 < n < 4 ⇒ n = - 3, -2, -1, 0, 1, 2, 3
Now, 2√2 sin
is integer for n = 0, 3, - 3
2√2 sin
is irrational for n = - 2, -1, 1, 2
⇒ cos(3α - 7β) sin (5β - α)
= sin(3α - 7β) cos(5β - α)
⇒ sin[(5β - α) - (3α - 7β)] = 0
⇒ sin(4α - 12β) - (3α - 7β)] = 0
⇒ sin(4α - 12β) = 0
⇒ 4(α - 3β) = nπ ⇒
∵ α - 3β ∈ (-π, π)
⇒
⇒ - 4 < n < 4 ⇒ n = - 3, -2, -1, 0, 1, 2, 3
Now, 2√2 sin
is integer for n = 0, 3, - 32√2 sin
is irrational for n = - 2, -1, 1, 2Create a free account to view solution
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