IBDP Maths SL 3.8 Solving trigonometric equations AA HL Paper 1- Exam Style Questions- New Syllabus

Question

Solve the equation \( 3\sin^2x + 4\cos x = 1 \), for \( 0 \leq x \leq 2\pi \).

▶️ Answer/Explanation

Solution

Use the identity \( \sin^2x = 1 – \cos^2x \).

\( 3(1 – \cos^2x) + 4\cos x = 1 \)

\( 3 – 3\cos^2x + 4\cos x = 1 \)

Rearrange:
\( 3\cos^2x – 4\cos x – 2 = 0 \)

Let \( y = \cos x \), solve:
\( 3y^2 – 4y – 2 = 0 \)

Using the quadratic formula:
\[ y = \frac{4 \pm \sqrt{(-4)^2 – 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 24}}{6} = \frac{4 \pm \sqrt{40}}{6} = \frac{4 \pm 2\sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3} \]

Evaluate:
\( \cos x = \frac{2 + \sqrt{10}}{3} \approx 1.387 \) (not valid since \( \cos x \leq 1 \))
\( \cos x = \frac{2 – \sqrt{10}}{3} \approx -0.387 \)

Now solve \( \cos x = -0.387 \) for \( 0 \leq x \leq 2\pi \)

\( x = \cos^{-1}(-0.387) \approx 1.97 \, \text{rad} \)

Since cosine is negative in Q2 and Q3:
\( x \approx 1.97 \) and \( x \approx 2\pi – 1.97 = 4.31 \)

✅ Final Answer: \( x \approx 1.97 \, \text{rad}, \, 4.31 \, \text{rad} \)

Question:

Solve \(\tan(2x – 5^\circ) = 1\) for \(0^\circ \leq x \leq 180^\circ\).

▶️ Answer/Explanation

Solution:

Step 1: Solve for the General Solution

The general solution for \(\tan\theta = 1\) is:

\[ \theta = 45^\circ + k180^\circ, \quad k \in \mathbb{Z} \]

Setting \(\theta = 2x – 5^\circ\), we get:

\[ 2x – 5^\circ = 45^\circ + k180^\circ \]

Step 2: Solve for \(x\)

Rearrange for \(x\):

\[ 2x = 50^\circ + k180^\circ \]

\[ x = \frac{50^\circ + k180^\circ}{2} \]

Step 3: Find Solutions in the Given Range

For \(k = 0\):

\[ x = \frac{50^\circ}{2} = 25^\circ \]

For \(k = 1\):

\[ x = \frac{50^\circ + 180^\circ}{2} = \frac{230^\circ}{2} = 115^\circ \]

For \(k = 2\):

\[ x = \frac{50^\circ + 360^\circ}{2} = \frac{410^\circ}{2} = 205^\circ \]

Since \(205^\circ\) is outside the given range \(0^\circ \leq x \leq 180^\circ\), we discard this solution.

Final Answer:

\[ x = 25^\circ, 115^\circ \]

Markscheme:

Attempt to equate \(2x – 5^\circ\) to reference angle. (M1)

\[ 2x – 5^\circ = 45^\circ, 225^\circ \] A1

\[ x = 25^\circ, 115^\circ \] A1

[3 marks]

Question:

Find the least positive value of \(x\) for which \(\cos\left(\frac{x}{2} + \frac{\pi}{3}\right) = \frac{1}{\sqrt{2}}\).

▶️ Answer/Explanation

\[\cos\left(\frac{x}{2} + \frac{\pi}{3}\right) = \frac{1}{\sqrt{2}}\]

\[\frac{x}{2} + \frac{\pi}{3} = \frac{\pi}{4} \Rightarrow x = -\frac{\pi}{6} \text{ (rejected)}\]

\[\frac{x}{2} + \frac{\pi}{3} = \frac{7\pi}{4}\]

\[x = \frac{17\pi}{6}\]

Markscheme:

Determines \(\frac{\pi}{4}\) as reference angle (A1)

Attempts to solve \(\frac{x}{2} + \frac{\pi}{3} = \frac{\pi}{4}, \frac{7\pi}{4}\) (M1)

Correct solution \(x = \frac{17\pi}{6}\) (A1)

[3 marks]

Question

Show that sin \(\frac{sinxtanx}{1-cosx}= 1+\frac{1}{cosx}\) x ≠ 2nπ, n ∈ R . [3]
Hence determine the range of values of k for which \(\frac{sinxtanx}{1-cosx}\)= K has no real solutions. [4]

▶️Answer/Explanation

Ans:

(a)

METHOD 1