Trigonometric equation $2\sin x+\cos x+1=0$

Question

I have to calculate $\dfrac{d}{dx}\dfrac{1+\cos x}{2+\sin x}=0$.

I have already simplified to: $2\sin x+\cos x+1=0$, but I have no idea how to go further..

Could someone give a hint?

Answer 1

Hint: $$2\sin(x)+\cos(x) \equiv \sqrt{5} \sin(x+a),$$ where $a=\arctan(1/2).$

Then we have $$\sqrt{5} \sin(x+a)=-1,$$ which you should be able to solve.

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You may be asking: "Where did the first identity come from?".

Answer: We can write $$\alpha\sin(x)+\beta\cos(x)$$ in the form $A \sin(x+\phi)$ for some $\phi$ and $A$.

i.e. $$\alpha\sin(x)+\beta\cos(x) \equiv A\sin(x+\phi).$$

Let's expand the RHS using the addition identity for sine.

$$\alpha\sin(x)+\beta\cos(x) \equiv A\underbrace{[\sin(x)\cos(\phi)+\cos(x)\sin(\phi)]}_{\equiv \ \sin(x+\phi)}.$$

Simplifying, we have $$\color{red}\alpha\sin(x)+\color{green}\beta\cos(x) \equiv \color{red}{A\cos(\phi)} \sin(x)+\color{green}{A\sin(\phi)}\sin(x).$$

This is an identity, so it's true for all (permitted) values of $x$.

Comparing coefficients of $\sin(x)$ we have $$\alpha=A\cos(\phi) \tag{1}.$$

Comparing coefficients of $\cos(x)$ we have $$\beta=A\sin(\phi) \tag{2}.$$

If we do $\frac{(2)}{(1)},$ we have $$\frac{A\sin(\phi)}{A\cos(\phi)}=\Large \boxed{\tan(\phi)=\frac{\beta}{\alpha}}.$$

If we do $(1)^2+(2)^2$, we have $[A\cos(\phi)]^2+[A\sin(\phi)]^2=A^2[\sin^2(\phi)+\cos^2(\phi)]=A^2(1)=A^2=\alpha^2+\beta^2,$ from which we get $$\Large\boxed{A=\sqrt{\alpha^2+\beta^2}}$$

Use the boxed equations in your example, with $\alpha=2$ and $\beta=1$ to determine $A$ and $\phi$ (which I've told you is $\sqrt{5}$ and $\arctan(1/2)$ respectively, so try it if you don't believe me!).

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In general, if you've got an equation of the form $$\alpha \sin(x) +\beta \cos(x)=\gamma,$$ just transform the given equation into one of the form $$A\sin(x+\phi),$$ where $A$ and $\phi$ satisfy the boxed relationships (i.e. use the boxed formulae as opposed to deriving the whole think like I have).

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Note: you can express it in the form $A \cos(x+\phi)$ (if you don't like sine), but the boxed equations will be slightly different. Try it yourself and see what you come up with!

Answer 2

HINT:

I believe the linked Question is : Range of a trigonometric function

Using Double Angle formula, we have $$2\cdot2\sin\frac x2\cos\frac x2+2\cos^2\frac x2=0$$

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Alternatively apply Weierstrass Substitution to form a Quadratic Equation in $\tan\dfrac x2$

Answer 3

The solution of alexqwx gives you one solution $x_1=-\arctan(1/2)$. The other solution from the second answer is $\cos(x/2)=0$,i.e., $x_2=\pi$. Let $f(x)=\frac{1+\cos x}{2+\sin x}$, then $f(x_2)=\min(f(x))=0$, $f(x_1)=\max(f(x))=4/3.$

Here is a plot of $f(x)$ vs $x$ (the blue line is 4/3):