(2sin^2 x - sinx - 1)/(1 + cos2x)=0
2sin²x-sinx-1=0 или 1+cos2X=0
sinx=t cos2x=-1
2t²-t-1=0 2x=п+2пn
t=-1/2 x=п/2+пn
t=1
sinx=1
x=п/2+2пn
sinx=-1/2
x=((-1)^k+1)п/6+пk
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
2sin²x-sinx-1=0 или 1+cos2X=0
sinx=t cos2x=-1
2t²-t-1=0 2x=п+2пn
t=-1/2 x=п/2+пn
t=1
sinx=1
x=п/2+2пn
sinx=-1/2
x=((-1)^k+1)п/6+пk