Solve the equation " 2sec^2(x) = 5tanx " for 0 < x < π

Firstly, we must recognise that the equation contains two different trigonometric functions (sec() and tan()) and therefore we must rewrite one of these functions in terms of the other. Therefore we will need our trigonometric identity that contains both sec() and tan(): " sec^2(x) = tan^2(x) + 1 ". So by substituting this into our original equation gives us: " 2tan^2(x) + 2 = 5tan(x) " Now we can see that we have a quadratic equation in tan(x). To make this clearer I will say "Let tan(x) be represented by t", which gives: " 2t^2 - 5t + 2 = 0 " Now, like in GCSE we can factorise this equation to find its roots, like so: " (2t - 1)(t - 2) = 0 " which gives us our solution: " t = 1/2, t = 2 ". But we haven't finished yet as the question wants us to find values for x (in radians) which satisfy the equation given. To do this, we must replace t for tan(x). " tan(x) = 1/2, tan(x) = 2 " so our solution is: " x = arctan(1/2), x = arctan(2) " which approximately equals: " x = 0.464, x = 1.107 ".