Question

please help in this question ​

Answer 1

Answer:

well is the blue mark a correction to the question or just a random mark

Answer 2

$\frac{ {cos}^{2} x - 3cosx + 2}{ {sin}^{2} x} = 1 \\ {cos}^{2} x - 3cosx + 2 = {sin}^{2} x \\ {cos}^{2} x - 3cosx + 2 = 1 - {cos}^{2} x \: \: \: \: \: \: ...( {sin}^{2} x + {cos}^{2} x = 1) \\ 2 {cos}^{2} x - 3cosx + 1 = 0 \\ 2 {cos}^{2} x - 2cosx - cosx + 1 = 0 \\ 2cosx(cosx - 1) - 1(cosx - 1) = 0 \\ (2cosx - 1)(cosx - 1) = 0 \\ \\ 2cosx - 1 = 0 \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: \: cosx - 1 = 0 \\ cosx = \frac{1}{2} \: or \: 1 \\ cosx = cos {60}^{o} \: or \: cos {0}^{o} \\ x = {60}^{o} \: or \: {0}^{o} \\ \\ but \: x \: cannot \: be \: equal \: to \: zero \: since \: in \: the \: expression \: \\ {sin}^{2} x \: is \: in \: the \: denominator. \\ when \: x = 0 \\ sinx = 0 \: therefore \: {sin}^{2} x = 0 \\ division \: of \: anything \: by \: zero \: is \: not \: defined \\ \\ hence \\ x \: = {60}^{o}$