Trigonometric EquationHard
Question
General solution of the equation
3√3 sin3x + cos3x + 3√3 sin x cos x = 1 is -
3√3 sin3x + cos3x + 3√3 sin x cos x = 1 is -
Options
A.nπ + (-1)n
; n ∈ I
; n ∈ IB.2nπ, n ∈ I
C.nπ + (-1)n 
- 
- D.None of these
Solution
3√3 sin3x + cos3x + 3√ sin x cos x = 1
We know that
if a + b + c = 0 ⇒ a3 + b3 + c3 = 3abc
Now a = √3 sin x, b = cos x, c = -1
a3 + b3 + c3 - 3abc = 0 Possible only if a + b + c = 0
√3 sin x + cos x - 1 = 0 ⇒ √3 sin x + cos x = 1
sin x + 
cosx = 
⇒ sin 
= sin
⇒ x +
= nπ + (-1)n π/6 ⇒ x = nπ + (-1)n 
-
We know that
if a + b + c = 0 ⇒ a3 + b3 + c3 = 3abc
Now a = √3 sin x, b = cos x, c = -1
a3 + b3 + c3 - 3abc = 0 Possible only if a + b + c = 0
√3 sin x + cos x - 1 = 0 ⇒ √3 sin x + cos x = 1
sin x + cosx = ⇒ sin = sin ⇒ x +
= nπ + (-1)n π/6 ⇒ x = nπ + (-1)n -Create a free account to view solution
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