The curve $C$ has the equation $2\sin x + \cos y =1$, where $-\pi < x < \pi$ and $- \pi < y < \pi$.
- find an expression for $dy/dx$
- find the coordinates of the stationary points on $C$.
I get $(2\cos x -1)/( \sin y ) = dy/dx$ but I'm not sure if the derivative is only supposed to be in terms of $x$.
Please help.
The derivative of $ y' $ can be in terms of both $x $ and $y.$
In your case, let $ y = y(x)$ which gives $ 2\sin(x) + \cos(y(x)) = 1$. Since both sides are equal, you can derivate both sides and the equally will still hold. Since $ \frac{d}{dx} (1) = 0 $ you obtain $$ 2 \cos(x) - y'(x) \sin(y(x)) = 0 \ \leftrightarrow \ y'(x) = \frac{2cos(x)}{sin(y(x))}, \ y(x) \neq n \cdot\pi, \ n \in \mathbb{N}$$
Now (2) shouldn't be too complicated, given basic knowledge in calculus.