已知6sin²a+cosasina-2cos²a=0,a∈[π/2,π]，求sin(2a+π/3) 要很详细
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(3sina+2cosa)*(2sina-cosa)=0，∵a∈[π/2,π]，∴3sina+2cosa = 0，tana = -2/3.
∴ cosa = -1/√1+tan²a = -3/√13,sina = 2/√13
而 sin(2a+π/3) = sin2acosπ/3+cos2asinπ/3 = sinacosa + √3/2(2cos²a-1) = -6/13 + √3/2 *5/13
= (5√3-12)/26.
∴ cosa = -1/√1+tan²a = -3/√13,sina = 2/√13
而 sin(2a+π/3) = sin2acosπ/3+cos2asinπ/3 = sinacosa + √3/2(2cos²a-1) = -6/13 + √3/2 *5/13
= (5√3-12)/26.
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(3sina+2cosa)*(2sina-cosa)=0，∵a∈[π/2,π]，∴3sina+2cosa = 0，tana = -2/3.
∴ cosa = -1/√1+tan²a = -3/√13,sina = 2/√13
而 sin(2a+π/3) = sin2acosπ/3+cos2asinπ/3 = sinacosa + √3/2(2cos²a-1) = -6/13 + √3/2 *5/13
= (5√3-12)/26.
∴ cosa = -1/√1+tan²a = -3/√13,sina = 2/√13
而 sin(2a+π/3) = sin2acosπ/3+cos2asinπ/3 = sinacosa + √3/2(2cos²a-1) = -6/13 + √3/2 *5/13
= (5√3-12)/26.
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