Решение
cos^4x-sin^4x = √2 / 2
(cos²x)² - (sin²x)² = √2 / 2
(cos²x + sin²x)*(cos²x - sin²x) = √2 / 2
cos2x = √2/2
2x = (+ -)arccos(√2/2) + 2πk, k ∈ Z
2x = (+ -)*(π/4) + 2πk, k ∈ Z
x = (+ -)*(π/8) + πk, k ∈ Z
Решение
cos^4x-sin^4x = √2 / 2
(cos²x)² - (sin²x)² = √2 / 2
(cos²x + sin²x)*(cos²x - sin²x) = √2 / 2
cos2x = √2/2
2x = (+ -)arccos(√2/2) + 2πk, k ∈ Z
2x = (+ -)*(π/4) + 2πk, k ∈ Z
x = (+ -)*(π/8) + πk, k ∈ Z