# Trouble solving these trigonometric simultaneous equations

I have the following pair of simultaneous equations:

$$\sin(2x) + \sqrt3 \cos(2x) = -1$$ $$\sqrt3 \sin(2x) - \cos(2x) = \sqrt3$$

I need to answer the question: "what is the sum of the solutions to these equations, given that $$0?".

After some manipulation using identities, I get:

$$\cos(2x) = -\frac{\sqrt3}{2}$$

Which gives me solutions of:

$$75^o, 105^o, 255^o, 285^o$$

The sum of which is $$720^o$$

However, the answer to the question is $$330^o$$, because I should have been left with $$75^o$$ and $$255^o$$. I plugged the equations into Desmos and sure enough, this is what should be the case.

What have I done wrong? What I did was replace both sin(2x) terms with the identity 2sin(x)cos(x), eliminating this term and rearranging. Not sure what I'm missing here, aside from just manually plugging in all solutions and seeing if they satisfy both equations, but that feels like a bit of a manual fix, so to speak.

• Did the question state the range of acceptable solutions? Feb 13, 2021 at 13:49
• Not sure if this helps or not, but the bottom equation looks like the derivative of the top equation Feb 14, 2021 at 5:19
• @YvesDaoust it seems likely that the OP means in the range $0\le x \le 360^\circ$ Feb 14, 2021 at 12:17
• @YvesDaoust Yes, sorry, I knew I forgot to mention something! The solutions must be in the range 0<x<360. Feb 14, 2021 at 12:47
• You don't need identities. Just multiply the second equation by $\sqrt 3$ and add the result to the first to obtain that $\sin(2x)=0.5,$ which implies that $\sin(2x)-\sin(30°)=0,$ or that $$2\cos(x+15°)\sin(x-15°)=0.$$ This yields $x+15°=\pm 90°$ or $x-15°=0°$ or $x-15°=180°.$ Similarly, you can find that $$\cos(2x)=-\sqrt 3/2,$$ so that $2x$ must be in the second quadrant. Thus, $x$ must be such that $45°\le x\le 90°.$ Feb 14, 2021 at 20:28

For $$x=105^{\circ}$$ and $$x=285^{\circ}$$, $$\sin(2x)=-\frac{1}{2}$$ instead of $$\frac{1}{2}$$ so the solutions are $$75^{\circ}$$ and $$255^{\circ}$$ and their sum is $$330^{\circ}$$
• I don't really think that there is a way to do that.Another way that you can use which is closer to what you want is to find the interval where $2x$ is between $[0^{\circ},90^{\circ})$,$[90^{\circ},180^{\circ})$,$[180^{\circ},270^{\circ})$ and $[270^{\circ},360^{\circ})$ based on the signs of $cos(2x)$ and $sin(2x)$.In this way you can eliminate the wrong solutions without having to check anything,but from my own experience,I would recommend to always check the solutions that you got,especially when you are at a contest because you can lose points even though you know how to solve the problem Feb 14, 2021 at 16:36