Let A = Then must be one of the solutions to: Use identity Use identity on Use identity on Use identity on Use identity on Simplify: Factor sin²(A) out of the last two terms on the left: Use the identity on sin²(A) FOIL out the last term on the left: Let cos(A)=x Try a root of 1 1 | 16 0 -20 0 5 -1 | 16 16 -4 -4 1 16 16 -4 -4 1 0 So it factors as We try factoring as the product of two quadratics. In fact we might expect that it is the square of a quadratic because the leading term is a square and also the constant term 1 is a square. So we try to see if there is a quadratic (4x²+kx-1) which when squared gives Equating coefficients of x³: 8k = 16 k = 2 Equating coefficients of x²: k²-8 = -4 k² = 4 k = ±2 So far, k=2 works Equating coefficients of x -2k=-4 k=2 So indeed factors as So we have solutions x-1 = 0 and x = 1 So the solutions are , , . Since is in Q2, its cosine must be negative. The only solution which is negative is . thus . Edwin