Find an equivalent equation involving trig functions

[nickek]

Homework Statement

Given the equation $$\sin(2x)+2\sin(x)+2\cos(x) = 0$$ find an equivalent equation on the form $A+\sin(kx + v)=0$ (i.e find the values of the parameters A, k and v).

Relevant Equations

$$\sin(2x) = 2\sin(x)\cos(x)$$ $$2\sin(x)+2\cos(x) = 2\sqrt{2}\sin(x + \pi/4)$$

Rewrite the given equation, attempt 1:
$2\sin(x)\cos(x) + 2\sin(x) + 2\cos(x) = 0$
$\sin(x)\cos(x) + \sin(x) + \cos(x) = 0$
$\sin(x)(\cos(x) + 1) + \cos(x) = 0$, naaah, can't get any relevant out from here.

Attempt 2:
$2\sin(x)\cos(x) + 2\sqrt{2}*\sin(x + \pi/4) = 0$
$\sin(x)\cos(x) + \sqrt{2}\sin(x + \pi/4) = 0$, nope, don't know how continue.

I know I can rewrite $\cos(x)$ as $\sin(\pi/2 - x)$, but I don't see how it should help.

 

[fresh_42]

Homework Statement:: Given the equation sin(2x)+2*sin(x)+2*cos(x), find an equivalent equation on the form A+sin(kx + v)=0 (i.e find the values of the parameters A, k and v).
Relevant Equations:: sin(2x) = 2*sin(x)*cos(x), 2*sin(x)+2*cos(x) = 2*sqrt(2)*sin(x + pi/4).

Rewrite the given equation, attempt 1:
2*sin(x)*cos(x) + 2*sin(x) + 2*cos(x) = 0
sin(x)*cos(x) + sin(x) + cos(x) = 0
sin(x)*(cos(x) + 1) + cos(x) = 0, naaah, can't get any relevant out from here.

Attemt 2:
2*sin(x)*cos(x) + 2*sqrt(2)*sin(x + pi/4) = 0
sin(x)*cos(x) + sqrt(2)*sin(x + pi/4) = 0, nope, don't know how continue.

I know I can rewrite cos(x) as sin(pi/2 - x), but I don't see how it should help.

$\sin(2x) +2\sin(x)+2\cos(x)$ isn't an equation. And as a function, it is not of the form $A+\sin(kx+v)$ as you can see here:
https://www.wolframalpha.com/input?i=sin(x)cos(x)+sin(x)+cos(x)+=

Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

 

[nickek]

$\sin(2x) +2\sin(x)+2\cos(x)$ isn't an equation. And as a function, it is not of the form $A+\sin(kx+v)$ as you can see here:
https://www.wolframalpha.com/input?i=sin(x)cos(x)+sin(x)+cos(x)+=

Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

Sorry, I missed the equal sign in the problem statement and have now rewritten the post with mathematical typesetting.

 

[fresh_42]

You could try $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$ to find the zeros.
Or use the Weierstraß substitution $t=\tan(x/2)$ for $|x|<\pi.$

If we look at the solution

then there is an asymmetry which seems as if there will be two different solutions for the triple $(A,k,v).$

 

[Mark44]

Attempt 2: $2\sin(x)\cos(x) + 2\sqrt{2}*\sin(x + \pi/4) = 0$

This looks like a good start to me, with this change.
$\sin(2x) + 2\sqrt{2}*\sin(x + \pi/4) = 0$

You didn't show how you got from $2\sin(x) + 2\cos(x)$ to $2\sqrt 2\sin(x + \pi/4)$, but I assume that you know what you did and where the factor of $2\sqrt 2$ came from.

If you can convert $A\sin(2x) + B\sin(x + \pi/4)$ into an expression like $C\sin(kx + v)$ using the same technique that produced $2\sqrt 2 \sin(x + \pi/4)$, that should do it for you. It's possible that the constant A is zero, but I haven't worked things out.

BTW, it seems to me that you really aren't working with an equation, but rather, rewriting the expression involving sin(2x), sin(x), and cos(x) into an identically equal expression. But then again, I don't know what the exact wording of the problem is.