<s>[INST] Find the derivative of $y=\cos(x) - 2\sin(x),$ when the gradient is $1$ I need to find the smallest positive value of $x$ for which the gradient of the curve has value 1.
For this equation: 
$$
y =\cos(x)-2\sin(x)
$$
The answer is 2.5c grad.
The following is my working out but I’ve stopped at the factor formula, how can I change the equation having just $\sin()$ functions? Or is there another neat way?
$$
y = \cos(x)-2\sin(x) \\ 
\frac{d}{dx} \cos(x)-2\sin(x) = 1 \\ 
\frac{dy}{dx} =-\sin(x)-2\cos(x) = 1 \\ 
\frac{dy}{dx} =\sin(x)+2\cos(x) = -1 \\  
\frac{dy}{dx} = \sin^2(x)+2\cos(x)\sin(x)=-\sin(x)
$$
 [/INST] $$\sin x + 2\cos x + 1 = 0\tag{1}$$  Now use the Weierstrass substitution $y = \tan\left(\frac x2\right).$  From this, it follows that $$\sin x = \frac {2y}{y^2 + 1},\;\cos x = \frac{1-y^2}{y^2 + 1}\tag{2}$$
Then, substituting the values from $(2)$ into $(1)$ gives us $$ \frac {2y}{y^2 + 1} + \frac{2(1-y^2)}{y^2 + 1} + 1=0$$
Now, solving for $y$, $$\begin{align} &\quad \frac {2y}{y^2 + 1} + \frac{2(1-y^2)}{y^2 + 1} + 1=0 \\ \\
&\iff \dfrac{2y+2(1-y^2)+ y^2 + 1}{y^2 + 1} = 0\\ \\
&\iff \frac{y^2 - 2y - 3}{y^2 +1} = 0\\ \\
&\iff y^2 - 2y - 3 = (y - 3)(y+1) = 0\\ \\
& \iff y = 3\;\text{ or }\;y = -1
\end{align}$$
Now simply recall that $\;y = \tan\left(\frac x2\right) \implies \tan^{-1}(y) = \frac x2 \iff x = 2\tan^{-1}(y).$
So use $x = 2\tan^{-1}(y)$ to solve for $x$. One of your two $y$ values will yield the desired conclusion.
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