6 is the value of x in this
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B = C → A = C - B
→ B = C - A
Use the Double Angle Identity: cos 2A = 2 cos² A - 1
→ (cos 2A + 1)/2 = cos² A
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]
Use Even/Odd Identity: cos (-A) = cos (A)
<u>Proof LHS → RHS:</u>
LHS: cos² A + cos² B + cos² C
![\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201%2B%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A%2B2B%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2B%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%2B%5Ccos%5E2%20C%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1%2B%5Ccos%20%28A%2BB%29%5Ccdot%20%5Ccos%20%28A-B%29%2B%5Ccos%5E2%20C)
![\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%201%2B%5Ccos%20C%5B%5Ccos%20%28A-B%29%2B%5Ccos%20C%5D)
![\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201%2B%5Ccos%20C%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%2BC%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B-C%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1%2B2%5Ccos%20C%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2B%28C-B%29%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B-B-%28C-A%29%7D%7B2%7D%5Cbigg%29)
LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C 
Answer:
I dont have idea what I need to the to help you
Step-by-step explanation:
Answer:
x = (-19)/3
Step-by-step explanation:
Solve for x:
10 - 2 x + 5 (x + 4) + 3 (2 x + 9) = 0
5 (x + 4) = 5 x + 20:
10 - 2 x + 5 x + 20 + 3 (2 x + 9) = 0
3 (2 x + 9) = 6 x + 27:
6 x + 27 + 5 x - 2 x + 10 + 20 = 0
Grouping like terms, 6 x + 5 x - 2 x + 10 + 20 + 27 = (-2 x + 5 x + 6 x) + (10 + 20 + 27):
(-2 x + 5 x + 6 x) + (10 + 20 + 27) = 0
-2 x + 5 x + 6 x = 9 x:
9 x + (10 + 20 + 27) = 0
10 + 20 + 27 = 57:
9 x + 57 = 0
Subtract 57 from both sides:
9 x + (57 - 57) = -57
57 - 57 = 0:
9 x = -57
Divide both sides of 9 x = -57 by 9:
(9 x)/9 = (-57)/9
9/9 = 1:
x = (-57)/9
The gcd of 57 and 9 is 3, so (-57)/9 = (-(3×19))/(3×3) = 3/3×(-19)/3 = (-19)/3:
Answer: x = (-19)/3
