-2cos^2x+2sin^2x при (-пи/24)

Question

-2cos^2x+2sin^2x
при (-пи/24)

Answer 1

2cos^2x-2sin2x+1=0.
2cos^2x - 4sinxcosx + 1 = 0
2cos^2x - 4sinxcosx + sin^2x + cos^2x = 0
sin^2x - 4sinxcosx + 3cos^2x = 0 /: cos^2x не рао. 0
tg^2x - 4tgx +3=0
(tgx - 1 )(tgx - 3 ) = 0

1) tgx = 1 ==> x = pi/4 +pik, k c Z
2) tgx = 3 ===> x = arctg(3) +pik, k c Z