Solve : (tan (2x – 15)° – 1 = 0), for values of x such that (0° leq x leq 360°).
Explanation
(tan (2x – 15)° – 1 = 0 implies tan (2x – 15)° = 1)
((2x – 15)° = tan^{-1} (1) = frac{n pi}{4}, n = 1, 2, …)
(2x – 15 = 45°, 225°, 405°, 585°, 0° leq x leq 360°)
(2x = 60°, 240°, 420°, 600°)
(x = 30°, 120°, 210°, 300°)