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Superme
@Superme
July 2022
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ПОЖАЛУЙСТА РЕШИТЕ СРОЧНО !!!!!!
2cos2x=-√3/2
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MadMax123
Sinx+2*(1-2sin²x)=3/2
sinx+2-4sin²x-3/2=0
4sin²x-sinx-0,5=0
sinx=a
4a²-a-0,5=0
D=1+8=9
a1=(1-3)/8=-1/4⇒sinx=-1/4⇒x=(-1)^n+1 *arcsin1/4 +πn
a2=(1+3)/8=1/2⇒sinx=1/2⇒x=(-1)^n *π/6+πn
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тппомртлмрлпмргле
2cos^2x - 2sin^2x + 4sinx + 1 = 02 - 2sin^2x - 2sin^2x + 4sinx +1 = 0-4sin^2x + 4sinx + 3 = 0 sinx = t, -4t^2 + 4t + 3 = 04t^2 - 4t - 3 = 0D= 16 + 48 = 64 = 8^2 t1 = 12/8 = 1,5 (посторонний корень) t2 = 0,5sinx = 0,5
x = (-1)^n*п/6 + пn, n принадлежит Z.
1 votes
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тппомртлмрлпмргле
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Answers & Comments
sinx+2-4sin²x-3/2=0
4sin²x-sinx-0,5=0
sinx=a
4a²-a-0,5=0
D=1+8=9
a1=(1-3)/8=-1/4⇒sinx=-1/4⇒x=(-1)^n+1 *arcsin1/4 +πn
a2=(1+3)/8=1/2⇒sinx=1/2⇒x=(-1)^n *π/6+πn