Trigonometric EquationHard
Question
The number of intrgral values of k for which the equation 7cos x +5sin x = 2k +1 has asolution, is
Options
A.4
B.8
C.10
D.12
Solution
We know 
≤ a sin x + b cos x ≤ 
∴
≤ 7cos x + 5sin x 
ie, ≤ 2k + 1
Since, k is integer, - 9 < 2k + 1 < 9
⇒ -10 < 2k < 8
⇒ -5 < k < 4
⇒ Number of possible integer values of k = 8
≤ a sin x + b cos x ≤ ∴
≤ 7cos x + 5sin x ie,
≤ 2k + 1 Since, k is integer, - 9 < 2k + 1 < 9
⇒ -10 < 2k < 8
⇒ -5 < k < 4
⇒ Number of possible integer values of k = 8
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