The number of integer triplets $(a,b,c)$ such that $$a+b\cos(2x)+c\sin^2x=0\quad\text{for all $x$}$$ is
- (A) $0$
- (B) $1$
- (C) $3$
- (D) infinitely many
I tried to break $\cos 2x$ and write it in sine form, but got confused.
The number of integer triplets $(a,b,c)$ such that $$a+b\cos(2x)+c\sin^2x=0\quad\text{for all $x$}$$ is
I tried to break $\cos 2x$ and write it in sine form, but got confused.
Hint: $\cos{2x} = 1-2\sin^2{x}$.