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MrHUPPY
@MrHUPPY
July 2022
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2sin^x+3cosx-3=0
[4pi;5pi]
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Armenia2780
2sin²x+3cosx-3=0
2(1-cos²x)+3cosx-3=0
-2cos²x+3cosx-1=0
2cos²x-3cosx+1=0
cosx=t€[-1;1]
2t²-3t+1=0
D=9-8=1
t=(3±1)/4
t1=1;t2=1/2
1)cosx=1
x=2πn
2)cosx=1/2
x=±arccos1/2+2πn
x=±π/3+2πn;n€Z
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Answers & Comments
2(1-cos²x)+3cosx-3=0
-2cos²x+3cosx-1=0
2cos²x-3cosx+1=0
cosx=t€[-1;1]
2t²-3t+1=0
D=9-8=1
t=(3±1)/4
t1=1;t2=1/2
1)cosx=1
x=2πn
2)cosx=1/2
x=±arccos1/2+2πn
x=±π/3+2πn;n€Z