Using this format, we use
cos^2 x or (cosx)^2 to show the square of cosx
the ^ symbol is used to show exponents.
4 sin^2 x - 8cosx - 8 = 0
4(1 - cos^2 x) - 8cosx - 8 = 0
4 - 4cos^2 x - 8cosx - 8 = 0
4cos^2 x + 8cosx + 4 = 0
cos^2 x + 2cosxs + 1 = 0
(cosx + 1)^2 = 0
cosx + 1 = 0
cosx = -1
in the first period x = π
so for your domain
x = π + 2π = 3π
Find all solutions of 4(sin(x)**2)-8cos (x) -8 = 0 in the interval (2pi, 4pi).
(Leave your answers in exact form and enter them as a comma-separated list.)
1 answer