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If$f(x) = {\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x$, then number of values of $x \in [0,2\pi ]$for which $f(x) = 0$ are

Last updated date: 17th Jul 2024
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Hint: All the terms in the right hand side are squared. Think about what we can deduce from here.

 Given, $f(x) = {\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x$.
Since, every term in the right-hand side of the equation is squared, the value of each individual term could be either zero or greater than zero. We are interested in $x$ where $f(x) = 0$. For$f(x)$to be zero, each individual on the right-hand side has to be zero. That is${\cos ^2}x = 0,{\cos ^2}2x = 0{\text{ and }}{\cos ^2}3x = 0$.Now let’s solve them one by one.
\[{\cos ^2}x = 0 \Leftrightarrow \cos x = 0 \Leftrightarrow x = \frac{\pi }{2},\frac{{3\pi }}{2}\]
\[{\cos ^2}2x = 0 \Leftrightarrow \cos 2x = 0 \Leftrightarrow 2x = \frac{\pi }{2},\frac{{3\pi }}{2} \Leftrightarrow x = \frac{\pi }{4},\frac{{3\pi }}{4}\]
\[{\cos ^2}3x = 0 \Leftrightarrow \cos 3x = 0 \Leftrightarrow 3x = \frac{\pi }{2},\frac{{3\pi }}{2} \Leftrightarrow x = \frac{\pi }{6},\frac{{3\pi }}{6}\]
Observe that, there is no common value of $x$ in all the above terms. The question should come in our mind as to why we are finding the common values. It’s just because we want $x$ where$f(x) = 0$ and $f(x) = 0$ when all the individual terms on the right hand side will be zero. It means for a single value of $x$, all the terms on the right-hand side has to vanish simultaneously. That’s why we are looking at the common value of $x$ where ${\cos ^2}x = 0,{\cos ^2}2x = 0{\text{ and }}{\cos ^2}3x = 0$.But, there is no such common value in the given domain. So, there is no $x$ for which $f(x) = 0$.
Hence the correct option is D

Note: When you are finding the roots of something, keep domain in your mind. Here we have given our domain as $f(x) = 0$. So, we only considered such x where $f(x) = 0$ in the given domain. One should not step out of the domain.