If cos4x - sin4x = \(\frac{3}{5}\), find 1 - 2sin2x + 2 sinx cosx is equal to:

\(\frac{3}{4}\) \(\frac{7}{5}\) \(\frac{5}{7}\) 0

\(\frac{7}{5}\)

⇒ cos4x - sin4x = \(\frac{3}{5}\) ⇒ (cos2x + sin2x) (cos2x - sin2x) = \(\frac{3}{5}\) ⇒ (cos2x - sin2x) = \(\frac{3}{5}\) ⇒ (cos 2A) = \(\frac{3}{5}\) So, ⇒ (1 - 2sin2x) + (2 sinx cosx) = (cos2x) + (sin2x) = \(\frac{3}{5}\) + \(\frac{4}{5}\) = \(\frac{7}{5}\)