How do you solve this equation?

3^2+4^2-2(3)(4)cos(x)=5^2+6^2-2(5)(6)cos(180-x)

9+16-24cos(x)=25+36-60cos(180-x)

25-24cos(x)=61-60cos(180-x)

*cos(180-x)=-cos(x)*

-36-24cos(x)=-60(-cos(x))

-36-24cos(x)=60cos(x)

-36=84cos(x)

-36/84=cos(x)

-3/7=cos(x)

Since cosine is negative in both quadrants II and III,

x=115.4 and x=295.4

3^2 + 4^2 − 2(3)(4)cos(x) = 5^2 + 6^2 − 2(5)(6)cos(180 − x)

∴ 9 + 16 − 24cos(x) = 25 + 36 − 60cos(180 − x)

∴ 60cos(180 − x) − 24cos(x) = 36

∴ 12(5cos(180 − x) − 2cos(x)) = 36

∴ 5cos(180 − x) − 2cos(x) = 3

given cos(α − β) = cos(α)cos(β) + sin(α)sin(β)

∴ 5cos(180)cos(x) + 5sin(180)sin(x) − 2cos(x) = 3

Strictly speaking as written that's 180 radians not 180° but I'll assume you mean 180° as it give a clean answer

∴ -5cos(x) − 2cos(x) = 3

∴ -7cos(x) = 3

∴ cos(x) = -3/7

∴ x = 360°n ± cos⁻¹(-3/7) where n∈ℤ