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