by 1st equation  sec y = √(1+cos²x)  so cos y = 1/ √(1+ cos²x)

by 3rd equation sec z = cot x  so  tan z = √(cot²x - 1)

on putting these 2 values in 2nd equation we get  

1/ √(1+ cos²x) = √(cot²x - 1)

so sin²x = (1+cos²x)(cos²x-sin²x)

or sin²x = (2-sin²x)(1-2sin²x)

or sin²x = 2-4sin²x-sin²x+2sin⁴x

or 2sin⁴x-6sin²x+2 = 0

now solve it for sinx