A guide curve in a design plot, drawn in scaled coordinates, has the slope dy/dx = 1/(x cos y + sin…

Step 1. Turn it over. $\dfrac{dx}{dy} - (\cos y)\,x = \sin 2y$. Here $P_1 = -\cos y$, so the integrating factor is $e^{-\sin y}$.

Step 2. Integrate. $x\,e^{-\sin y} = \displaystyle\int 2\sin y\cos y\,e^{-\sin y}\,dy$. Put $u = \sin y$, so $du = \cos y\,dy$. The integral is $\int 2u\,e^{-u}\,du = -2(u + 1)e^{-u} + K$.

Step 3. Divide by the factor. $x = -2(\sin y + 1) + Ke^{\sin y}$. Here $K$ is the constant $C$ of the designer's form.

Step 4. The constant. At $(0, 0)$: $0 = -2 + K$, so $C = 2$. The curve is $x = 2e^{\sin y} - 2\sin y - 2$.

Step 5. Evaluate. At $y = \dfrac{\pi}{2}$, $\sin y = 1$, so $x = 2e - 2 - 2 = 1.4366$. At $y = \dfrac{\pi}{6}$, $\sin y = 0.5$, so $x = 2e^{0.5} - 1 - 2 = 0.2974$.