Ecuatia cos^{4}x-sin^{4}x=1+sin2x are multimea de solutii

(cos²x-sin²x) (cos²x+sin²x)=1+sin2x

cos²x-sin²x=cos²x+sin²x+sin2x

-sin²x=sin²x+2sinxcosx

2sinxcosx+2sin²x=0

2sinx(cosx+sinx) =0

sinx=0 ..x=kπ

cosx+sinx=0 |:cosx≠0, x≠(2k+1)π/2

1+tgx=0

tgx=-1

x=kπ +arctg(-1) =kπ-π/4

Deci

S= kπ∪{kπ-π/4}, k∈Z

verificare

pt x=kπ

1+0=1+0 adevarat, bine rezolvat

pt x=-π/4

(√2/2)^4-(√2/2)^4=1+sin (-π/2)

0=1-1

adevarat, bine rezolvat