I believe that for this question you will use the quadratic equation which is: (-b +/- sqrt(b^2-4(a)(c))/2a Plug in -5=a, 39=b, 22=c and that should give you 2 answers for t.
Explanation:
To achive that you have to piant a ring in red, which will have its big diameter equal to the sphere diameter and its small diameter equal to the cube diagonal.
The diagonal of the cube can be calculated using Pithagoras:
![D^2=L^2+L^2](https://tex.z-dn.net/?f=D%5E2%3DL%5E2%2BL%5E2)
Where D is the diagonal and L is the side of the cube
Answer:
11h+13
Step-by-step explanation:
7h+4h=11h
9+4=13
ANSWER
![x = \frac{\pi}{2} , \frac{7\pi}{6} , \frac{3\pi}{2} , \frac{11\pi}{6}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%2C%20%5Cfrac%7B7%5Cpi%7D%7B6%7D%20%2C%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%2C%20%5Cfrac%7B11%5Cpi%7D%7B6%7D%20)
EXPLANATION
The given trigonometric equation is
![\cos(x) + 2 \cos(x) \sin(x) = 0](https://tex.z-dn.net/?f=%20%5Ccos%28x%29%20%20%2B%202%20%5Ccos%28x%29%20%20%5Csin%28x%29%20%20%3D%200)
We factor cos(x) to get:
![\cos(x) (1 + 2 \sin(x) ) = 0](https://tex.z-dn.net/?f=%20%5Ccos%28x%29%20%281%20%2B%202%20%5Csin%28x%29%20%29%20%3D%200)
Apply the zero product property to obtain:
![\cos(x) = 0 \: or \: 1 + 2 \sin(x) = 0](https://tex.z-dn.net/?f=%20%5Ccos%28x%29%20%20%3D%200%20%5C%3A%20or%20%5C%3A%201%20%2B%202%20%5Csin%28x%29%20%20%3D%20%200)
![\cos(x) = 0 \: or \: \sin(x) = - \frac{1}{2}](https://tex.z-dn.net/?f=%20%5Ccos%28x%29%20%20%3D%200%20%5C%3A%20or%20%5C%3A%20%5Csin%28x%29%20%20%3D%20%20%20-%20%20%5Cfrac%7B1%7D%7B2%7D%20)
Using the unit circle,
![\cos(x) = 0](https://tex.z-dn.net/?f=%20%5Ccos%28x%29%20%20%3D%200)
when
![x = \frac{\pi}{2}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20)
and
![x = \frac{3\pi}{2}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20)
We know
![\sin(y) = \frac{1}{2}](https://tex.z-dn.net/?f=%20%5Csin%28y%29%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20)
when
![y= \frac{\pi}{6}](https://tex.z-dn.net/?f=y%3D%20%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20)
The sine function is negative in the third and fourth quadrants.
![x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}](https://tex.z-dn.net/?f=x%20%3D%20%5Cpi%20%2B%20%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%20%3D%20%20%5Cfrac%7B7%5Cpi%7D%7B6%7D%20)
![x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}](https://tex.z-dn.net/?f=x%20%3D%202%5Cpi%20%20-%20%20%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%20%3D%20%20%5Cfrac%7B11%5Cpi%7D%7B6%7D%20)
Hence the solutions are:
![x = \frac{\pi}{2} , \frac{7\pi}{6} , \frac{3\pi}{2} , \frac{11\pi}{6}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%2C%20%5Cfrac%7B7%5Cpi%7D%7B6%7D%20%2C%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%2C%20%5Cfrac%7B11%5Cpi%7D%7B6%7D%20)
3(x+2) = 2(2x+7)
3x + 6 = 4x + 14 (On the left side we distribute 3 through parenthesis and on the right side we do the same with 2)
3x - 4x = 14 - 6 (Here we move 4x from the right to the left side and it changes its sign. We do the same with 6)
-x = 8
x = - 8