for \(0^\circ\le x\le360^\circ\), write down the value of \(k\).
Solution
Answer: The graph is a sine curve with midline \(y=-2\), maximum \(1\), minimum \(-5\), and one complete cycle from \(0^\circ\) to \(360^\circ\). Also, \(k=5\).
For \(y=3\sin x-2\), the midline is \(y=-2\) and the amplitude is \(3\).
So the maximum value is
\(-2+3=1,\)
and the minimum value is
\(-2-3=-5.\)
The curve starts at \((0^\circ,-2)\), reaches a maximum at \((90^\circ,1)\), returns to \((180^\circ,-2)\), reaches a minimum at \((270^\circ,-5)\), and returns to \((360^\circ,-2)\).
For part (ii), the expression \(3\sin x-2\) ranges from \(-5\) to \(1\). Therefore \(|3\sin x-2|\) ranges from \(0\) up to the largest absolute value, which is
