Question 1 Report
Solve : \(\tan (2x - 15)° - 1 = 0\), for values of x such that \(0° \leq x \leq 360°\).
Solve \(\tan(2x-15)^{\circ}-1=0\) for \(0^{\circ}\le x\le 360^{\circ}\).
The tangent is \(1\) at \(45^{\circ}\) and every \(180^{\circ}\) thereafter, so
Take values of \(x\) in \([0^{\circ},360^{\circ}]\):
Hence \(x=30^{\circ},\ 120^{\circ},\ 210^{\circ},\ 300^{\circ}\).
Answer Details
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