Solve $5 \sin x - 4 = 2 \cos 2x$ for $0 \leq x < 2\pi$. - Scottish Highers Maths - Question 6 - 2017

Rearranging the equation yields:

$4 \sin^2 x + 5 \sin x - 6 = 0$

Factoring gives:

$(4\sin x + 3)(\sin x - 2) = 0$

Setting each factor to zero, we have:

1. ( 4\sin x + 3 = 0 ) which leads to ( \sin x = -\frac{3}{4} )
2. ( \sin x - 2 = 0 ) which suggests ( \sin x = 2 ), but this has no solution within the range.

Thus, we focus on ( \sin x = -\frac{3}{4} ).

The solutions within ( 0 \leq x < 2\pi ) are determined using the arcsine function:

( x = \arcsin(-\frac{3}{4}) ) gives us two angles in the third and fourth quadrants. Specifically:

1. ( x \approx 3.67 ) radians
2. ( x \approx 5.54 ) radians