Trigonometrical Equations 3 Question 8
9. The number of all possible triplets $\left(a _1, a _2, a _3\right)$ such that $a _1+a _2 \cos (2 x)+a _3 \sin ^{2}(x)=0, \forall x$ is
(1987, 2M)
(a) 0
(b) 1
(c) 3
(d) $\infty$
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Answer:
Correct Answer: 9. (d)
Solution:
- Given, $\quad a _1+a _2 \cos 2 x+a _3 \sin ^{2} x=0, \forall x$
$\Rightarrow a _1+a _2 \cos 2 x+a _3 \frac{1-\cos 2 x}{2}=0, \forall x$
$\Rightarrow \quad a _1+\frac{a _3}{2}+a _2-\frac{a _3}{2} \cos 2 x=0, \forall x$
$\Rightarrow \quad a _1+\frac{a _3}{2}=0 \quad$ and $\quad a _2-\frac{a _3}{2}=0$
$\Rightarrow \quad a _1=-\frac{k}{2}, a _2=\frac{k}{2}, a _3=k$, where $k \in R$
Hence, the solutions, are $-\frac{k}{2}, \frac{k}{2}, k$, where $k$ is any real number.
Thus, the number of triplets is infinite.
