When the angle is a compound argument such as 2x-30, do not solve for x straight away. Solve for the whole argument first — over its own adjusted interval, because as x runs through the stated interval the argument runs through a different one. - Adjust the interval: apply to every part of the inequality the same operations that build the argument. - Find the reference angle and choose the quadrants as usual. - Keep adding or subtracting 360 (or 2pi) until every solution inside the adjusted interval has been found. - Undo the operations to return to x, and check each answer against the original interval. **Adjusting the interval for the argument 2x-30 ** **Step** | **Working** Stated interval | 0 <= x <= 360 Multiply every part by 2 | 0 <= 2x <= 720 Subtract 30 from every part | -30 <= 2x-30 <= 690 **Solving sin(2x-30)= 12 for 0 <= x <= 360 ** **Step** | **Working** Reference angle | alpha=sin-1 12=30 Quadrants I and II (sine positive) | 2x-30 =30, 150 Add 360 within -30 <= 2x-30 <= 690 | 2x-30 =30, 150, 390, 510 Return to x | 2x=60, 180, 420, 540 x=30, 90, 210, 270 **Twice the turns, twice the solutions** Over one turn of x, the argument 2x completes two turns, so sin 2x=k collects twice as many solutions as sin x=k on the same interval. Adjust the interval before solving, and only undo the operations at the very end. Converting back to x too early — or forgetting the extra +- 360 rounds inside the adjusted interval — loses solutions.

Equations with Compound Angles

Secondary 4

TGM Original Questions

When the angle is a compound argument such as \(2x-30^\circ\), do not solve for \(x\) straight away. Solve for the whole argument first — over its own adjusted interval, because as \(x\) runs through the stated interval the argument runs through a different one.

Method

Adjust the interval: apply to every part of the inequality the same operations that build the argument.
Find the reference angle and choose the quadrants as usual.
Keep adding or subtracting \(360^\circ\) (or \(2\pi\)) until every solution inside the adjusted interval has been found.
Undo the operations to return to \(x\), and check each answer against the original interval.

Adjusting the interval for the argument \(2x-30^\circ\)

Step	Working
Stated interval
Multiply every part by \(2\)
Subtract \(30^\circ\) from every part

Solving \(\sin(2x-30^\circ)=\dfrac12\) for \(0^\circ\le x\le 360^\circ\)

Step	Working
Reference angle
Quadrants I and II (sine positive)
Add \(360^\circ\) within \(-30^\circ\le 2x-30^\circ\le 690^\circ\)
Return to \(x\)

Twice the turns, twice the solutions

Over one turn of \(x\), the argument \(2x\) completes two turns, so \(\sin 2x=k\) collects twice as many solutions as \(\sin x=k\) on the same interval.

Exam check

Adjust the interval before solving, and only undo the operations at the very end. Converting back to \(x\) too early — or forgetting the extra \(\pm 360^\circ\) rounds inside the adjusted interval — loses solutions.

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