Trig eqn cos(x+TT/6) + cos(x-TT/4) = 0

Hi there, I was wondering if someone would be able to help me solve this equation :

cos(x+TT/6) + cos(x-TT/4) = 0

It's driving me insane. I use compound angle theorem to expand the components and end up with:

$\displaystyle \sqrt{3}/2\cos{x}-1/2\sin{x}+1/\sqrt{2}cos{x}+1/\sqrt{2}sin{x}=0$

I then tried squaring both sides and using trig identities to put everything in terms of $\displaystyle cos^2{x}$. And got something that I wont waste your time with. Please help!

Solution by exploiting symmetry

Quote:

Originally Posted by

**flashylightsmeow** $\displaystyle \cos\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0$

By $\displaystyle \cos\left(\pi+\theta \right) = -\cos{\theta}$, we have $\displaystyle \cos\left(x-\frac{\pi}{4}\right) = \cos\left(\pi+x-\frac{\pi}{4}-\pi\right) = -\cos\left(x-\frac{5\pi}{4}\right)$.

Thus $\displaystyle \cos\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0 \Rightarrow \cos\left(x+\frac{\pi}{6}\right) = -\cos\left(x-\frac{\pi}{4}\right) = \cos\left(x-\frac{5\pi}{4}\right).$

So $\displaystyle \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right)$. If $\displaystyle \cos{\theta} = \cos{\alpha}$, then $\displaystyle \theta = 2n\pi + \alpha$ or $\displaystyle \theta = 2n\pi -\alpha$, thus:

$\displaystyle \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi + x - \frac{5\pi}{4} \Rightarrow 0 = 2n\pi-\frac{17\pi}{12}{\pi}~ $ (dead-end).

Or we have $\displaystyle \cos\left(x+\frac{\pi}{6}\right) = \cos\left(x-\frac{5\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi -x + \frac{5\pi}{4} \Rightarrow x = n\pi+\frac{13\pi}{24} ~ \checkmark. $

Most problems of this kind can be solved by some consideration of symmetry; a twin of your problem is:

Quote:

$\displaystyle \sin\left(x+\frac{\pi}{6}\right) +\cos\left(x-\frac{\pi}{4}\right) = 0$

$\displaystyle \cos{\theta} = \sin\left(\frac{\pi}{2}-\theta\right)$, thus $\displaystyle \cos\left(x-\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}+\frac{\pi}{4}-x\right) = \sin\left(\frac{3\pi}{4}-x\right)$, hence

$\displaystyle \sin\left(x+\frac{\pi}{6}\right) = -\cos\left(x-\frac{\pi}{4}\right) \Rightarrow \sin\left(x+\frac{\pi}{6}\right) = -\sin\left(\frac{3\pi}{4}-x\right) = \sin\left(x-\frac{3\pi}{4}\right).$

So $\displaystyle \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right)$. If $\displaystyle \sin{\theta} = \sin{\alpha}$, then $\displaystyle \theta = 2n\pi+\alpha$ or $\displaystyle (2n+1)\pi-\alpha$, thus

$\displaystyle \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = 2n\pi+x-\frac{3\pi}{4} \Rightarrow 0 = 2n\pi-\frac{3\pi}{4}-\frac{\pi}{6} $ (dead-end).

Or $\displaystyle \sin\left(x+\frac{\pi}{6}\right) = \sin\left(x-\frac{3\pi}{4}\right) \Rightarrow x+\frac{\pi}{6} = (2n+1)\pi-x+\frac{3\pi}{4} \Rightarrow x = (2n+1)\frac{\pi}{2}+\frac{7\pi}{24} ~ \checkmark. $