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If $f(x) = \left( {\dfrac{{\sin 3x}}{{\sin x}}} \right),x \ne n\pi ,$ then the range of values of $f(x)$ for real values of $x$ is
(a)$[1 - 3)$
(b)$( - \infty , - 1]$
(c)$(3, + \infty )$
(d)$[ - 1,3]$

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Last updated date: 03rd Mar 2024
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IVSAT 2024
Answer
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Hint:As we know that the above given question is related to trigonometric expression, sine and cosine are trigonometric ratios. Here we have to find the value of $f(x)$, so first of all we have to solve and simplify the value. We can convert the equation into basic trigonometric equations by applying the trigonometric identities.

Complete step by step solution:
As per the given question we have to solve the expression $f(x) = \dfrac{{\sin 3x}}{{\sin x}}$. We know that a trigonometric identity for $\sin 3x$ which is $3\sin x - 4{\sin ^3}x$, now by substituting this value we will expand and we get, $f(x) = \dfrac{{3\sin x - 4{{\sin }^3}x}}{{\sin x}}$. We can take the common $\sin x$ out and it gives,
$\dfrac{{\sin x(3 - 4{{\sin }^2}x)}}{{\sin x}} = 3 - 4{\sin ^2}x$. Now let us assume $3 - 4{\sin ^2}x = y$, so we get: ${\sin ^2}x = \dfrac{{3 - y}}{4}$.
Since we can say that $0 \leqslant \dfrac{{3 - y}}{4} \leqslant 1$. Now solving this by inequality :
$ = 0 \leqslant 3 - y \leqslant 4 \Rightarrow - 3 \leqslant - y \leqslant 1$.
Now we can interchange the values but by keeping the signs not changed, $ - 1 \leqslant y \leqslant 3$.
So we can write it as $y \in [ - 1,3]$.
Hence the correct option is (d) $[ - 1,3]$.

Note: Before solving this kind of question we should have the proper knowledge of all trigonometric ratios, identities and their formulas. To solve this trigonometric expression we should also have the proper knowledge of the inequality, and then we should solve it by avoiding mistakes and taking care of positive and negative signs.
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