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dariaAppp
@dariaAppp
October 2021
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sedinalana
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[2(cosx+sinx)+2sin²x)]/2(1+sinx)=√3+sinx
2(cosx+sinx+sin²x)/2(1+sinx)=√3+sinx
cosx+sinx+sin²x=√3+sinx+√3sinx+sin²x
cosx=√3+√3sinx
cos²x/2-sin²x/2=√3(1+sinx)
(cosx/2-sinx/2)(cosx/2+sinx/2)=√3(sin²x/2+cos²x/2+2sinx/2cosx/2)
(cosx/2-sinx/2)(cosx/2+sinx/2)-√3(cosx/2+sinx/2)²=0
(cosx/2+sinx/2)(cosx/2-sinx/2-√3cosx/2-√3sinx/2)=0
(cosx/2+sinx/2)[cosx/2(1-√3)-sinx/2(1+√3)]=0
cosx/2+sinx/2=0/cosx/2
1+tgx/2=0
tgx/2=-1
x/2=-π/4+πn
x=-π/2+2πn,n∈z
[cosx/2(1-√3)-sinx/2(1+√3)=0/cosx/2
(1-√3)-tgx/2(1+√3)=0
tgx/2=(1-√3)/(1+√3)
tgx/2=√3-2
x/2=-arctg(2-√3)+πk
x=-2arctg(2-√3)+2πk,k∈z
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Answers & Comments
Verified answer
[2(cosx+sinx)+2sin²x)]/2(1+sinx)=√3+sinx2(cosx+sinx+sin²x)/2(1+sinx)=√3+sinx
cosx+sinx+sin²x=√3+sinx+√3sinx+sin²x
cosx=√3+√3sinx
cos²x/2-sin²x/2=√3(1+sinx)
(cosx/2-sinx/2)(cosx/2+sinx/2)=√3(sin²x/2+cos²x/2+2sinx/2cosx/2)
(cosx/2-sinx/2)(cosx/2+sinx/2)-√3(cosx/2+sinx/2)²=0
(cosx/2+sinx/2)(cosx/2-sinx/2-√3cosx/2-√3sinx/2)=0
(cosx/2+sinx/2)[cosx/2(1-√3)-sinx/2(1+√3)]=0
cosx/2+sinx/2=0/cosx/2
1+tgx/2=0
tgx/2=-1
x/2=-π/4+πn
x=-π/2+2πn,n∈z
[cosx/2(1-√3)-sinx/2(1+√3)=0/cosx/2
(1-√3)-tgx/2(1+√3)=0
tgx/2=(1-√3)/(1+√3)
tgx/2=√3-2
x/2=-arctg(2-√3)+πk
x=-2arctg(2-√3)+2πk,k∈z