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 7 - 2019 - Paper 1

To solve the equation

$5 \sin 2\theta = 9 \tan \theta$

we start by rewriting it in terms of a single trigonometric function. Using the double angle identity for sine, we have:

$\sin 2\theta = 2 \sin \theta \cos \theta$

Substituting this in gives:

$5 \cdot 2 \sin \theta \cos \theta = 9 \tan \theta$

Next, recall that:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

so we can rewrite the tangent term:

$10 \sin \theta \cos \theta = 9 \cdot \frac{\sin \theta}{\cos \theta}$

Multiplying both sides by (\cos \theta):

$10 \sin \theta \cos^2 \theta = 9 \sin \theta$

Assuming (\sin \theta \neq 0), we can divide by (\sin \theta):

$10 \cos^2 \theta = 9$

Now, we can simplify this to:

$\cos^2 \theta = \frac{9}{10}$

Taking the square root gives:

$\cos \theta = \pm \sqrt{\frac{9}{10}} = \pm \frac{3}{\sqrt{10}}$

For the positive case, we find:

$\theta = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right) \approx 18.4°$

For the negative case:

$\theta = 180° - 18.4° = 161.6°$

Thus, the solutions for (θ) in the interval ([-180°, 180°]) are:

$\theta \approx 18.4°, 161.6°$

Checking for any additional angles, we only find:

$\theta = -180° + 18.4° = -161.6°$