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The period of the function 

f(x)=4 \sin ^4\left(\frac{4 x-3 \pi}{6 \pi^2}\right)+2 \cos \left(\frac{4 x-3 \pi}{3 \pi^2}\right) 

is 

Option: 1

\frac{3 \pi^2}{4}


Option: 2

\frac{3 \pi^3}{4}


Option: 3

\frac{4 \pi^2}{3}


Option: 4

\frac{4 \pi^3}{3}


Answers (1)

best_answer

\begin{aligned} \because f(x) & =\left\{2 \sin ^2\left(\frac{4 x-3 \pi}{6 \pi^2}\right)\right\}^2+2 \cos \left(\frac{4 x-3 \pi}{3 \pi^2}\right) \\ \\& =\left\{1-\cos \left(\frac{4 x-3 \pi}{3 \pi^2}\right)\right\}^2+2 \cos \left(\frac{4 x-3 \pi}{3 \pi^2}\right) \\ \\& =1+\cos ^2\left(\frac{4 x-3 \pi}{3 \pi^2}\right) \end{aligned}

                \begin{aligned} & =\frac{1}{2}\left\{2+1+\cos \left(\frac{8 x-6 \pi}{3 \pi^2}\right)\right\} \\ \\& =\frac{3}{2}+\frac{1}{2} \cos \left(\frac{8 x-6 \pi}{3 \pi^2}\right) \end{aligned}

\therefore Period of f(x)=\frac{2 \pi}{\left(\frac{8}{3 \pi^2}\right)}=\frac{6 \pi^3}{8}=\frac{3 \pi^3}{4}

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Kshitij

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