5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β

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5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β
5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β

5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β
5cos^2a+4cos^2β=4cosα
令cosα = t
5t^2+4cos^2β=4t
4cos^2β=4t-5t^2
4cos^2β≥0
4t-5t^2≥0
5t^2-4t≤0
t(5t-4)≤0
0≤t≤4/5
5cos^2a+4cos^2β=4cosα
4cos^2a+4cos^2β=4cosα-cos^2α=4-4+4cosα-cos^2α=4-(2-cosα)^2=4-(2-t)^2
0≤t≤4/5
0≥ -t ≥ -4/5
2 ≥ 2-t ≥ 6/5
4 ≥ (2-t)^2 ≥ 36/25
-4≤-(2-t)^2≤-36/25
0≤4-(2-t)^2≤64/25
即0≤4cos^2a+4cos^2β≤64/25
0≤cos^2a+cos^2β≤16/25

5cos^2a+4cos^2β=4cosα>=0
4cos^2a+4cos^2β=4cosα-cos^2a=-[cosa-2]^2+4>=0
0<=cosa<=1
故cos^2a+cos^2β=(-[cosa-2]^2+4)/4=1-[cosa-2]^2
易知范围是[0,3]

-5/4到3/4闭区间

5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β=
解析：∵5cos^2a+4cos^2β=4cosα>=0
4cos^2a+4cos^2β=4cosα-cos^2a=-[cosa-2]^2+4
∴cos^2a+cos^2β=-1/4[cosa-2]^2+1
∵0<=cosa<=1
令f(x)=1-1/4(x-2)^2 (0<=x<=1)
∴f(x)的值域，即cos^2a+cos^2β取值范围是[0,3/4]

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