Cos70cos10+sin70sin10 e =2cos^2 2x-1

Answer 1

Cos70cos10+sin70sin10 e =2cos^2 2x-1
don't know what that "e" is supposed to be, seems to be superfluous.

recall : cos (A-B) = cosAcoB + sinAsinB
Cos70cos10+sin70sin10 =2cos^2 2x-1
cos(70-10) = 2cos^2 2x - 1
cos60 = 2cos^2 2x - 1
recall cos 2A = 2cos^2 A - 1
so 2cos^2 2x - 1 = cos 4x

cos 4x = cos60
4x = 60° or 4x = 300°
x = 15° or x = 75°

since cos 2x has a period of 180°
other solutions would be
x = 15+180 = 195°
or
x = 75+180 = 255°