在三角形ABC已知(6sinA-2sinc)cosB+sin(C-B)=sinA,若b=2根号3,三角形的面积为3根号2
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(6sinA - 2sinC)cosB+sin(C-B) = sinA

6sinAcosB - 2sinCcosB +sinCcosB - cosCsinB = sinA

6sinAcosB - (sinCcosB+cosCsinB) = sinA

6sinAcosB - sin(C+B) = sinA

∵sin(C+B) = sin(π - A) = sinA

∴6sinAcosB = 2sinA

cosB = 1/3

sinB = 2√2 /3

S = 1/2 ac sinB = √2 /3 ac = 3√2

∴ac = 9

b² = a²+c² - 2accosB

12 = a²+c² - 2*9*1/3

a²+c² = 18

(a+c)² - 2ac =18

(a+c)² = 18+2ac = 36

a+c = 6

ac=9

∴a=c=3

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