Can you explain to me what is wrong with this question?

Mathematics

2 answers:

Citrus2011 [14]3 years ago

8 0

Answer:

The thing that is wrong with the question is you put two but you need to put the 2 on top and the 1 and 3 at the bottom.

Step-by-step explanation:

2 over 1- ✔ 3

Anni [7]3 years ago

6 0

Answer:

I'm pretty sure it would be negative square root of 3 over 1 or 2. Sorry if this isn't much help.

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Find the general solution of <img src="https://tex.z-dn.net/?f=2cos%282x%20%2B%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%29%20%3D%20%5Cs

Aleks04 [339]

Answer:

$\large\boxed{x=-\dfrac{\pi}{4}+k\pi\ \text{or}\ x=k\pi\ \text{for}\ k\in\mathbb{Z}}$

Step-by-step explanation:

$2\cos\left(2x+\dfrac{\pi}{4}\right)=\sqrt2\qquad\text{divide both sides by 2}\\\\\cos\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}\to2x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+2k\pi\\\\2x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+2k\pi\ \vee\ 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+2k\pi\qquad\text{subtract}\ \dfrac{\pi}{4}\ \text{from all sides}\\\\2x=-\dfrac{2\pi}{4}+2k\pi\ \vee\ 2x=2k\pi\qquad\text{divide all sides by 2}\\\\x=-\dfrac{\pi}{4}+k\pi\ \vee\ x=k\pi\ \text{for}\ k\in\mathbb{Z}$