Question

cos(4*x) = cos(3*x)^2+sin(x)^2

Answer 1

Recall the following identities:

cos(4x) = cos⁴(x) - 6 cos²(x) sin²(x) + sin⁴(x)

cos(3x) = cos³(x) - 3 cos²(x) sin(x)

sin²(x) = 1 - cos²(x)

Then the equation

cos(4x) = cos²(3x) + sin²(x)

can be rewritten entirely in terms of cos(x) as

8 cos⁴(x) - 8 cos²(x) + 1 = 16 cos⁶(x) - 24 cos⁴(x) + 8 cos²(x) + 1

16 cos⁶(x) - 32 cos⁴(x) + 16 cos²(x) = 0

16 cos²(x) (cos²(x) - 1)² = 0

Now we can solve:

cos²(x) = 0   or   (cos²(x) - 1)² = 0

cos(x) = 0   or   cos²(x) - 1 = 0

cos(x) = 0   or   cos²(x)  = 1

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

x = π/2 + 2nπ   or   x = -π/2 + 2nπ

…   or   x = π + 2nπ

…   or   x = 2nπ

where n is any integer.