Question

If $f\left(x\right)={\cos }^{2}x+{\cos }^{2}2x+{\cos }^{2}3x$, then the umber of values of $x\in [0,2\pi ]$ for which $f\left(x\right)=1$ is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

Answer: (D)

$10$

$f\left(x\right)=1$

${\cos }^{2}3x+{\cos }^{2}2x+co{s}^{2}x=1$

$\cos 3x=4{\cos }^{3}x-3\cos x$

$\cos 2x=2{\cos }^{2}x-1$

now put these values in above equation

$(4{\cos }^{3}x-3\cos x{)}^2$ + $(2{\cos }^{2}x-1{)}^2$ + ${\cos }^{2}x=1$

$16{\cos }^{6}x-20{\cos }^{4}x+6\cos x=0$

${\cos }^{2}x(8{\cos }^{4}x-10{\cos }^{2}x+3)=0$

${\cos }^{2}x=0$ or $(8{\cos }^{4}x-10{\cos }^{2}x+3)=0$

${\cos }^{2}x=\frac{10{+}_{-}\sqrt{100-96}}{16}$

so,

${\cos }^{2}x=0,\frac{1}{2},\frac{3}{4}$

so, each root of ${\cos }^{2}x$ contributes two values in x∈[0,2π]x∈[0,2π]

no. of solutions are $\frac{\pi }{6}$ , $\frac{5\pi }{6}$ ,$\frac{\pi }{2}$ , $\frac{\pi }{3}$ ,$\frac{3\pi }{2}$ ,$\frac{2\pi }{3}$$\frac{4\pi }{3}$ $\frac{5\pi }{3}$ $\frac{7\pi }{6}$ ,$\frac{11\pi }{6}$