Trigonometric EquationHard
Question
Total number of integral values of ′n′ so that sin x (sin x + cos x ) = n has at least one solution is -
Options
A.2
B.1
C.3
D.zero
Solution
sin x (sin x + cos x) = n
sin2x + sin x cos x = n
= n ⇒ sin2x - cos2x = 2n -1
⇒ - √2 ≤ 2n - 1 ≤ √2 ⇒
≤ n ≤ 
n = 0, 1
sin2x + sin x cos x = n
= n ⇒ sin2x - cos2x = 2n -1⇒ - √2 ≤ 2n - 1 ≤ √2 ⇒
≤ n ≤ n = 0, 1
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