4a-(2a+1)+(3-4a). найти корни уравнения. (2-x)^2-(x+3)(x-3)=3x

Question

Алгебра: 4a-(2a+1)+(3-4a). найти корни уравнения. (2-x)^2-(x+3)(x-3)=3x

stasikpavel · Accepted Answer

Sin2A = 2sinAcosA; cos2A = 2cos^2A - 1
sin3A = sin(A+2A) = sinAcos2A + cosAsin2A = sinA(2cos^2A-1) + cosA(2sinAcosA)
= 2sinAcos^2A - sinA + 2sinAcos^2A

cos3A = cos(A+2A) = cosAcos2A - sinAsin2A = cosA(2cos^2A-1) - sinA(2sinAcosA)
= 2cos^3A-cosA - 2sin^2AcosA

Hence the left side of your equation equals

(2sinAcosA+4sinAcos^2A) / (2cos^2A - 1 + 2cos^3A - 2sin^2AcosA), now replace sin^2A by 1-cos^2A

= (2sinAcosA+4sinAcos^2A) / (4cos^3A + 2cos^2A -2cosA - 1)
= 2sinAcosA(1+2cosA) / ((2cos^2A-1)(1+2cosA))
= 2sinAcosA / (2cos^2A - 1)
= sin2A / cos2A
= tan2A

MOGZ ответил