Note that since ABC is a right triangle, there must be an angle such that its cosine is 0. There are two cases.
1.cosB=0: This is a contradiction, since $4=cosA+cosC\leq2$
2.Let cosA=0:Then, we have 8cosB+cosC=4 where C=pi/2-B. Then, 8cosB+cos(pi/2-B)=4, i.e, 8cosB+sinB=4. Therefore, 64(1-sin^2B)=(4-sinB)^2.
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