(a) By using a suitable trigonometrical identity, solve the equation tan(2x-π/6)^2 =11-sec(2x-π/6)giving all values of x in radians to two decimal places in the interval 0<=x <=π .

say (2x-π/6)=qtan²q = sec²q-1so sec²q-1= 11 - sec q sec²q + sec q -12 = 0(sec q -3) (sec q + 4) = 0sec q = 3 or -4because 0<=x<= πso -π/6 <= q <= 11π/6q = 1.23 or 1.82 or 4.46 or 5.05so x = 0.88 or 1.17 or 2.49 or 2.79