IB Mathematics SL 3.8 Solving trigonometric equations AA SL Paper 2- Exam Style Questions- New Syllabus

(a)
Use the double-angle identity:

\(\cos 2x=1-2\sin^2x\)

Therefore:

\(3\cos 2x+11\sin x=3(1-2\sin^2x)+11\sin x\)

\(3\cos 2x+11\sin x=3-6\sin^2x+11\sin x\)

\(3\cos 2x+11\sin x=3+11\sin x-6\sin^2x\)

✅ Hence, \(3\cos 2x+11\sin x=3+11\sin x-6\sin^2x\).

(b)
Using the result from part (a):

\(3+11\sin x-6\sin^2x-6=0\)

\(-6\sin^2x+11\sin x-3=0\)

\(6\sin^2x-11\sin x+3=0\)

Factorize:

\((3\sin x-1)(2\sin x-3)=0\)

Therefore:

\(\sin x=\dfrac13\)

or

\(\sin x=\dfrac32\)

The equation \(\sin x=\dfrac32\) has no real solution because \(-1\leq\sin x\leq1\).

For \(\sin x=\dfrac13\):

\(x=\sin^{-1}\left(\dfrac13\right)=19.471\ldots^\circ\)

The second solution between \(0^\circ\) and \(180^\circ\) is:

\(x=180^\circ-19.471\ldots^\circ=160.528\ldots^\circ\)

✅ Answer: \(x=19.5^\circ\) or \(x=161^\circ\)