IB Mathematics SL 3.8 Solving trigonometric equations AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Solve \( \tan(2x – 5^\circ) = 1 \) for \( 0^\circ \leq x \leq 180^\circ \).
\[ \tan^{-1}(1) = 45^\circ \]
General solution: \( 2x – 5^\circ = 45^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z} \)
\[ 2x = 50^\circ + k \cdot 180^\circ \]
\[ x = \frac{50^\circ + k \cdot 180^\circ}{2} \]
For \( 0^\circ \leq x \leq 180^\circ \):
Case \( k = 0 \): \( x = \frac{50^\circ}{2} = 25^\circ \)
Case \( k = 1 \): \( x = \frac{50^\circ + 180^\circ}{2} = \frac{230^\circ}{2} = 115^\circ \)
Case \( k = 2 \): \( x = \frac{50^\circ + 360^\circ}{2} = 205^\circ \) (outside interval)
Answer: \( x = 25^\circ, 115^\circ \)
▶️ Answer/Explanation
Part (a)
Find an expression for \( m \) in terms of \( q \).
For \( f(x) = \sin(qx) \), the period is \( \frac{2\pi}{q} \). The graph shows \( x \)-intercepts at \( x = 0 \), \( 2m \), and \( 4m \), indicating one full period from \( x = 0 \) to \( x = 4m \).
\[ 4m = \frac{2\pi}{q} \]
\[ m = \frac{\pi}{2q} \]
Answer: \( m = \frac{\pi}{2q} \)
Part (b)
Sketch the graph of \( g(x) = 3 \sin\left(\frac{2qx}{3}\right) \) for \( 0 \leq x \leq 6m \).
Compare \( g(x) = 3 \sin\left(\frac{2qx}{3}\right) \) to \( f(x) = \sin(qx) \):
– Amplitude of \( g \) is 3 (vs. 1 for \( f \)).
– Period of \( g \): \( \frac{2\pi}{\frac{2q}{3}} = \frac{6\pi}{2q} = \frac{3\pi}{q} = 3 \cdot \frac{\pi}{q} \). Since \( \frac{\pi}{q} = 2m \) (from part (a)), period of \( g \) is \( 3 \cdot 2m = 6m \).
– Horizontal stretch factor: \( \frac{\text{period of } g}{\text{period of } f} = \frac{6m}{4m} = \frac{3}{2} \).
The graph of \( g \) has:
– Amplitude 3, period \( 6m \), \( x \)-intercepts at \( x = 0, 3m, 6m \).
– Key points: Maxima at \( x = \frac{3m}{2} \), \( \frac{9m}{2} \) (\( y = 3 \)); minima at \( x = \frac{9m}{2} \) (\( y = -3 \)).
Sketch on the same axes shows a sine curve with triple the amplitude of \( f \), stretched horizontally by \( \frac{3}{2} \), over \( [0, 6m] \).
Answer: Graph as shown above.