Mathematics - Application of Derivative Question with Solution | TestHub

Let $\theta =\frac{\pi }{18}\Rightarrow 6\theta =\frac{\pi }{3}\Rightarrow \cos 6\theta =\frac{1}{2}$.

$$\begin{array}{rl}& \Rightarrow 4{\cos }^{3}2\theta -3\cos 2\theta =\frac{1}{2}\\ & \Rightarrow 8{\left(2{\cos }^2\theta -1\right)}^3-6\left(2{\cos }^2\theta -1\right)=1\end{array}$$

Let $2\cos \theta =x$.

$$\begin{array}{rl}& \Rightarrow 8{\left(2\cdot \frac{x^2}{4}-1\right)}^3-6\left(2\cdot \frac{x^2}{4}-1\right)=1\\ & \Rightarrow {\left(x^2-2\right)}^3-3\left(x^2-2\right)=1\\ & \Rightarrow x^6-6x^4+9x^2-3=0\end{array}$$

Let $f(x)=x^6-6x^4+9x^2-3$.

Then $f^{\prime }(x)=6x^5-24x^3+18x=6x\left(x^4-4x^2+3\right)$.

Setting $f^{\prime }(x)=0$ gives $x=0,\pm 1,\pm \sqrt{3}$.