▶️ Answer/Explanation
Solution
Let \( u = \sin^2\theta \), so the equation becomes:
\( 4u^2 + 12u – 7 = 0 \)
Solve using the quadratic formula:
\( u = \frac{-12 \pm \sqrt{12^2 – 4 \cdot 4 \cdot (-7)}}{2 \cdot 4} = \frac{-12 \pm \sqrt{144 + 112}}{8} = \frac{-12 \pm \sqrt{256}}{8} = \frac{-12 \pm 16}{8} \)
\( u = \frac{4}{8} = 0.5 \quad \text{or} \quad u = \frac{-28}{8} = -3.5 \)
Since \( u = \sin^2\theta \geq 0 \), discard \( u = -3.5 \).
Thus: \( \sin^2\theta = 0.5 \Rightarrow \sin\theta = \pm \sqrt{0.5} = \pm \frac{\sqrt{2}}{2} \)
For \( 0^\circ \leq \theta \leq 360^\circ \):
– \( \sin\theta = \frac{\sqrt{2}}{2} \): \( \theta = 45^\circ, 135^\circ \)
– \( \sin\theta = -\frac{\sqrt{2}}{2} \): \( \theta = 225^\circ, 315^\circ \)
Answer: \( 45^\circ, 135^\circ, 225^\circ, 315^\circ \)
▶️ Answer/Explanation
Solution
Compute \( \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \).
The equation becomes:
\( \frac{\sqrt{3}}{2} + \tan 2x + \frac{\sqrt{3}}{2} = 0 \)
\( \tan 2x + \sqrt{3} = 0 \)
\( \tan 2x = -\sqrt{3} \)
General solution: \( 2x = -\frac{\pi}{3} + k\pi, \quad k \in \mathbb{Z} \)
\( x = -\frac{\pi}{6} + \frac{k\pi}{2} \)
For \( -\frac{\pi}{4} < x < \frac{\pi}{4} \) (i.e., \( -0.785 < x < 0.785 \)):
– For \( k = 0 \): \( x = -\frac{\pi}{6} \approx -0.523 \), which satisfies the interval.
– For \( k = 1 \): \( x = -\frac{\pi}{6} + \frac{\pi}{2} \approx 1.047 \), which is outside.
– For \( k = -1 \): \( x = -\frac{\pi}{6} – \frac{\pi}{2} \approx -2.094 \), which is outside.
Answer: \( x = -\frac{\pi}{6} \)
▶️ Answer/Explanation
Solution
Part (a): State the values of \( a \), \( b \), and \( c \)
The curve is \( y = a\sin(bx) + c \). From the diagram:
– Amplitude (\( a \)): The curve oscillates between \( y = -1 \) and \( y = 7 \), so the amplitude is \( \frac{7 – (-1)}{2} = 4 \). Thus, \( a = 4 \).
– Vertical shift (\( c \)): The midline is at \( y = \frac{7 + (-1)}{2} = 3 \). Thus, \( c = 3 \).
– Frequency (\( b \)): The period is the length of one complete cycle. Over \( 0 \leq x \leq 2\pi \), the curve completes two cycles (observable from the diagram). The period is \( \frac{2\pi}{2} = \pi \). Since period = \( \frac{2\pi}{b} \), we have \( \frac{2\pi}{b} = \pi \Rightarrow b = 2 \).
Answer: \( a = 4 \), \( b = 2 \), \( c = 3 \)
Part (b): Number of solutions in \( 0 \leq x \leq 2\pi \)
Using \( a = 4 \), \( b = 2 \), \( c = 3 \), the function is \( y = 4\sin(2x) + 3 \), ranging from \( -1 \) to \( 7 \) with period \( \pi \).
(i) \( 4\sin(2x) + 3 = 7 – x \)
Find intersections of \( y = 4\sin(2x) + 3 \) and \( y = 7 – x \). The line \( y = 7 – x \) ranges from \( y = 7 \) at \( x = 0 \) to \( y = 7 – 2\pi \approx 0.716 \) at \( x = 2\pi \). Graphically, the sine wave completes two cycles, and the line crosses it multiple times due to its gradual slope. Numerical or graphical analysis (as implied by the answer) suggests 5 intersections within \( 0 \leq x \leq 2\pi \).
Answer: 5 solutions
(ii) \( 4\sin(2x) + 3 = 2\pi(x – 1) \)
Find intersections of \( y = 4\sin(2x) + 3 \) and \( y = 2\pi(x – 1) \). The line ranges from \( y = -2\pi \approx -6.283 \) at \( x = 0 \) to \( y = 2\pi(\pi – 1) \approx 6.853 \) at \( x = \pi \). Due to the steep slope of the line and the oscillatory nature of the sine function, graphical analysis indicates only one intersection within \( 0 \leq x \leq 2\pi \).
Answer: 1 solution