Sin^x-cos^x=корень из2/2

Question

Sin^x-cos^x=корень из2/2

Answer 1

Cos^2x - sin^2x = - √2/2
cos2x = - √2/2
2x = (+ -)arccos(- √2/2) + 2πn, n∈Z
2x = (+ -)(π - arccos(√2/2)) + 2πn, n∈Z
2x = (+ -)(π - π/4) + 2πn, n∈Z
2x = (+ -)(3π/4) + 2πn, n∈Z
x = (+ -)(3π/8) + πn, n∈Z