Trigonometric EquationHard
Question
The most general values of x for which sin x + cos x = 
{1, a2 - 4a + 6} are given by-
{1, a2 - 4a + 6} are given by-Options
A.2n π
B.2n π + π/2
C.nπ + (-1)n 
D.All of above
Solution
Now a2 - 4a + 6 = (a - 2)2 + 2 > 1
So
{1, a2 - 4a + 6} = 1 ⇒ sin x + cos x = 1
= sin
x +
= nπ + (-1)n 
⇒ x = nπ + (-1)n 
So
{1, a2 - 4a + 6} = 1 ⇒ sin x + cos x = 1= sin
x +
= nπ + (-1)n ⇒ x = nπ + (-1)n Create a free account to view solution
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