It is given that:
![\begin{gathered} 2\cos 2x+\sqrt[]{2}=0 \\ 2\cos 2x=-\sqrt[]{2} \\ \cos 2x=-\frac{\sqrt[]{2}}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202%5Ccos%202x%2B%5Csqrt%5B%5D%7B2%7D%3D0%20%5C%5C%202%5Ccos%202x%3D-%5Csqrt%5B%5D%7B2%7D%20%5C%5C%20%5Ccos%202x%3D-%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%20%5Cend%7Bgathered%7D)
cos2x is negative square root 2 divided by 2 in the second and third quadrants, so it follows:

Here theta is given by:

So the solution is given by:
Answer: The answer is B......
Step-by-step explanation:
Answer:
<h2><u><em>3x + 8.75 Or 3x + ³⁵⁄₄</em></u></h2>
Explanation:
- I'm going to convert all the fractions into decimals
⁻¹⁄₂x + 9.5 + x - ¾ + 2.5x
- = -0.5x + 9.5 + x - 0.75 + 2.5x
-0.5x + x + 2.5x + 9.5 - 0.75
- Add similar elements: -0.5x + x + 2.5x = 3x
= 3x + 9.5 - 0.75
- Add / subtract: 9.5 - 0.75 = 8.75
<h2><u><em>3x + 8.75 Or 3x +³⁵⁄₄</em></u></h2>
- PNW