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# Find the equation of the tangent and normal to the curves.$\;y=2\sin^{2} 3x$ at $\;x=\large\frac{\pi}{6}$

Note: This is part 3 rd of a 4 part question, split as 4 separate questions here.

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) y=2 \sin ^2 3x at x=\large\frac{\pi}{6} Step 1: When x=\large\frac{\pi}{6},$$y=2 \sin ^2 \large\frac{\pi}{2}$$=2 We are required to find the equation of the tangent and normal at \bigg(\large\frac{\pi}{6}$$,2 \bigg)$
Step 2:
Slope of the tangent at $\bigg(\large\frac{\pi}{6},$$2\bigg)=\large\frac{dy}{dx_{(\Large\frac{\pi}{6},2)}} \large\frac{dy}{dx}$$=4 \sin 3x. 3 \cos 3x=12 \sin 3x \cos 3x=6 \sin 6x$
$m=\large\frac{dy}{dx_{(\Large\frac{\pi}{6},2)}}=$$6 \sin \large\frac{\pi}{n}=0 Therefore the tangent is || to the x axes and to equation is y=2 Step 3: The normal is \perp to the tangent and so is || to the axes . It passes through \bigg(\large\frac{\pi}{6}$$,2\bigg)$.
Therefore the equations of the normal is $x=\large\frac{\pi}{6}$