$\cfrac{{dy}}{{dx}} + 2y\tan x = \sin x;y = 0$ when $x = \cfrac{\pi }{3}$

: The given equation is $\cfrac{{dy}}{{dx}} + 2y\tan x = \sin x$

…(1)
which is a linear equation of the type $\cfrac{{dx}}{{dy}} + Px = Q$

Where, $P = 2\tan x$ and $Q = \sin x$

$\therefore \int {Pdx} = \int {2\tan xdx = 2\int {\tan xdx = 2\log \left| {\sec x} \right| = \log {{\sec }^2}x} }$

$\therefore {\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int {Pdx} }} = {e^{\log {{\sec }^2}x}} = {\sec ^2}x$

$\therefore$ The solution is $y \cdot {\rm{(I}}{\rm{.F}}{\rm{.)}} = \int {Q \cdot {\rm{(I}}{\rm{.F}}{\rm{.)}}dx} + C$

$\Rightarrow \;y{\sec ^2}x = \int {\sin } x{\sec ^2}xdx + C = \int {\sec x\tan xdx} + C$

$\Rightarrow \;y{\sec ^2}x = \sec x + C$
…(2)
When $x = \cfrac{\pi }{3},y = 0$; then $0 = \sec \cfrac{\pi }{3} + C \Rightarrow C = - 2$

Putting $C = - 2$ in (2),

we get $y{\sec ^2}x = \sec x - 2$
$\Rightarrow y = \cos x - 2{\cos ^2}x$

which is the required solution.