Show that the equation 2sin^2(x) + 3sin(x) = 2cos(2x) + 3 can be written as 6sin^2(x)+3sin(x) - 5 = 0. Hence solve for 0 < x < 360 degrees. Giving your answers to 1.d.p.

2sin^2(x) + 3sin(x) = 2cos(2x) + 32sin^2(x) + 3sin(x) = 2(cos^2(x) - sin^2(x) ) + 32sin^2(x) + 3sin(x) = 2(1- 2sin^2(x) ) + 32sin^2(x) + 3sin(x) = 2 - 4sin^2(x) + 36sin^2(x) + 3sin(x) - 5 = 0 let sin(x) = yThen solve : 6Y^2 + 3Y - 5 = 0Y = [-3(+ or - ) root(129) ] / 12 = sin(x)sin(x) result always between [-1,1]. As such one of these solutions can be disregarded as it lies outside of this range. Take arcsin of the other result.Gives x = 44.1 degrees. Second solution in the range is 135.9 degrees. Second result calculated from graph or ASTC circle.