Given that 9 sin^2y-2 sin y cos y=8 show that (tany - 4)(tany + 2)= 0

This question requires us to make use of the trigonometric identities tan(y)=sin(y)/cos(y) and sin^2y + cos^2y = 1 which are given in the formula sheet of the exam. Since we know that sin(y)/cos(y) can be substituted for tan(y), our objective is to create some form of sin(y)/cos(y) in the first equation to enable us to get to tan(y). Therefore, we divide the first equation by cos^2y. This gets us9sin^2y/cos^2y - 2sinycosy/cos^2y = 8/cos^2ySince sin(y)/cos(y)=tan(y), this means that sin^2(y)/cos^2(y) = tan^2(y). Therefore, we can simplify the above equation to 9tan^2y-2tany=8/cos^2y. Now we have the equation in tan(y) form but we need to get rid of the cos^2y denominator and further simplify the equation. Using the sin^2y+cos^2y=1 identity, we can replace 8 by 8(cos^2y+sin^2y). We can simplify 8(cos^2y+sin^2y)/cos^2y by cancelling out the top cos^2y and converting the sin^2y/cos^2y part into tan^2y. This gets us 9tan^2y-2tany=8+8tan^2y. We must remember to keep the the first 8 as we must multiply the 1 we simplified the cos^2y/cos^2y into by 8 as 8 is outside of the brackets. From here, we bring 8+8tan^2y to the left hand side by subtracting them both from the LHS. From doing this we get tan^2y-2tany-8=0 which we can factorise to get the final equation (tany-4)(tany+2)=0