Solve the equation \( \sin^2 x + 2 = 3 \cos 2x \) for \( 0 \leq x < 360 \). - Scottish Highers Maths - Question 7 - 2023

Rearranging the equation gives:

$\sin^2 x + 6\sin^2 x - 3 + 2 = 0$

Which simplifies to:

$7\sin^2 x - 1 = 0$

Now we can rearrange to solve for ( \sin^2 x ):

$7\sin^2 x = 1$

Thus,

$\sin^2 x = \frac{1}{7}$

Taking the square root gives:

$\sin x = \pm \sqrt{\frac{1}{7}}$

Or,

$\sin x = \frac{1}{\sqrt{7}} \quad \text{or} \quad \sin x = -\frac{1}{\sqrt{7}}$

To find values of ( x ), we look for:

1. For ( \sin x = \frac{1}{\sqrt{7}} ): ( x \approx 19.47^{\circ} ) and ( x \approx 160.52^{\circ} )

2. For ( \sin x = -\frac{1}{\sqrt{7}} ): ( x \approx 202.54^{\circ} ) and ( x \approx 339.48^{\circ} )

Final solutions: ( x \approx 19.47^{\circ}, 160.52^{\circ}, 202.54^{\circ}, 339.48^{\circ} )