根号3 /2 ×sin2x+1/2cos2x=3/5 怎么解cos2x得几啊
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根号3 /2 ×sin2x+1/2cos2x=3/5 怎么解cos2x得几啊
根号3 /2 ×sin2x+1/2cos2x=3/5 怎么解cos2x得几啊
根号3 /2 ×sin2x+1/2cos2x=3/5 怎么解cos2x得几啊
根号3 /2 ×sin2x+1/2cos2x=3/5
(√3/2)sin2x=3/5-(1/2)cos2x
两边平方
(3/4)(1-cos²2x)=9/25-(3/5)cos2x+(1/4)cos²2x
cos²2x-(3/5)cos2x-39/100=0
解得cos2x=(3±4√3)/10
√3 /2 ×sin2x+1/2cos2x=3/5
(√3 /2 ×sin2x+1/2cos2x)/1=3/5
(√3 /2 ×2sinxcosx+1/2cos^2x-1/2sin^2x)/(sin^2x+cos^2x)=3/5
(√3tanx+1/2-1/2tan^2x)/(1+tan^2x)=3/5
6(1+tan^2x)=5(2√3tanx+1-tan^...
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√3 /2 ×sin2x+1/2cos2x=3/5
(√3 /2 ×sin2x+1/2cos2x)/1=3/5
(√3 /2 ×2sinxcosx+1/2cos^2x-1/2sin^2x)/(sin^2x+cos^2x)=3/5
(√3tanx+1/2-1/2tan^2x)/(1+tan^2x)=3/5
6(1+tan^2x)=5(2√3tanx+1-tan^2x)
11tan^2x-10√3tanx+1=0
tanx=(10√3±16)/22=(5√3±8)/11
tan^2x=(139±80√3)/121
cos2x=(1-tan^2x)/(1+tan^2x)
cos2x=(-9-40√3)/(130+40√3)或cos2x=(-9+40√3)/(130-40√3)
怎么数值这么难?
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