Question

The number of integral triples (a, b, c) such that a + b cos 2x + c $si{n}^2$ = 0, for all x, is

• A. 0
• B. 1
• C. 3
• D. infinitely many

Answer 1

The correct option is D

infinitely many

We have a + b cos 2x + c $si{n}^2$x = 0, for all x

a + b + (c – 2b) $si{n}^2$x = 0 for all x

a + b = 0 and c – 2b = 0 $\Rightarrow$ a = –b and c = 2b

Thus, the triplets are (–b, b, 2b), where b R.

Hence, there are infinitely many triplets.