Question

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

• A. $4$
• B. $6$
• C. $8$
• D. $0$

Answer 1

Answer: (D)

$0$

Given $f\left(x\right)={\cos }^{2}x+{\cos }^{2}2x+{\cos }^{2}3x$

Since every term in $f\left(x\right)$ is always $\ge 0$ . Therefore for $f\left(x\right)$ to be zero , each term should be zero

$\Rightarrow {\cos }^{2}x=0,{\cos }^{2}2x=0,{\cos }^{2}3x=0$

For above equation , there are no common solutions

Therefore the number of values of $x$ for which $f\left(x\right)=0$ is $0$