4cos^2(x)-7cos(x)+3
Please help and explain how you did it :)

I assume you're asked to solve

4 cos²(x) - 7 cos(x) + 3 = 0

Factor the left side:

(4 cos(x) - 3) (cos(x) - 1) = 0

Then either

4 cos(x) - 3 = 0   or   cos(x) - 1 = 0

cos(x) = 3/4   or   cos(x) = 1

From the first case, we get

x = cos⁻¹(3/4) + 2nπ  or   x = -cos⁻¹(3/4) + 2nπ

and from the second,

x = nπ

where n is any integer.

236 cm 16 time 12 = 192

6 times 7 = 42

192+42