If y = y(x) is the solution of the differential equation, (dy)/(dx)...

$\dfrac{dy}{dx} + 2ytanx = \sin x$

$I . F . = {e}^{∫ 2tanxdx} = {e}^{2\ln \sec x}$

$I . F . = \sec^{2}x$

$y . \left(\sec^{2}x\right) = ∫ \sin x . \sec^{2}xdx + C$

$y . \left(\sec^{2}x\right) = ∫ \sec x \tan xdx + C$

$y . \left(\sec^{2}x\right) = \sec x + C$

$x = \dfrac{\pi}{3}; y = 0$

$⇒ C = - 2$

$⇒ y = \dfrac{\sec x - 2}{\sec^{2}x} = \cos x - 2\cos^{2}x$

Let $\cos x = t, - 1 \le t \le 1$

$⇒ y = t - 2{t}^{2} ⇒ \dfrac{dy}{dt} = 1 - 4t = 0 ⇒ t = \dfrac{1}{4}$

Second-order derivative is negative

$∴ max = \dfrac{1}{4} - \dfrac{1}{8} = \dfrac{2 - 1}{8} = \dfrac{1}{8}$