6. (a) Solve, for -180° ≤ θ ≤ 180°, the equation 5 sin 2θ = 9 tan θ giving your answers, where necessary, to one decimal place - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 1

To solve the equation, start with the equation:

$5 sin 2θ = 9 tan θ$

Using the identity $sin 2θ = 2 sin θ cos θ$, we can rewrite the equation:

$5(2 sin θ cos θ) = 9 \frac{sin θ}{cos θ}$

This simplifies to:

$10 sin θ cos θ = 9 sin θ$

Assuming $sin θ ≠ 0$, we can divide both sides by $sin θ$:

$10 cos θ = 9$

Now we find:

$cos θ = \frac{9}{10}$

To find θ, calculate:

$θ = \arccos \left(\frac{9}{10}\right) ≈ 25.84°$

Also, since cosine is positive in the first and fourth quadrants, we will have another solution:

$θ = -25.84° + 360° \times k, k \in \mathbb{Z}$

Applying the range -180° ≤ θ ≤ 180° gives:

-$θ ≈ 25.8°$ -$θ ≈ -25.8°$

Thus, the solutions in the specified range, rounded to one decimal place, are:

• $θ ≈ 25.8°$
• $θ ≈ -25.8°$