Given the expression: \( \frac{\sin 150^{\circ}+\cos ^{2} x-1}{2} \) \( 5.3 .1 \quad \begin{array}{l}\text { Without using a calculator, simplify the expression given above to a } \\ \text { single trigonometric term in terms of } \cos 2 x\end{array} \) 5.3 .2 Hence, determine the general solution of \( \frac{\sin 150^{\circ}+\cos ^{2} x-1}{2}=\frac{1}{25} \) (6)

To simplify the expression \( \frac{\sin 150^{\circ}+\cos^2 x-1}{2} \), we start by noting that \( \sin 150^{\circ} = \frac{1}{2} \). So the expression becomes: \[ \frac{\frac{1}{2} + \cos^2 x - 1}{2} = \frac{\cos^2 x - \frac{1}{2}}{2}. \] Next, we can recognize \( \cos^2 x = \frac{1 + \cos 2x}{2} \). Substituting this in gives: \[ \frac{\frac{1 + \cos 2x}{2} - \frac{1}{2}}{2} = \frac{\frac{\cos 2x}{2}}{2} = \frac{\cos 2x}{4}. \] Thus, the simplified expression is \( \frac{\cos 2x}{4} \). Now, to solve the equation \( \frac{\cos 2x}{4} = \frac{1}{25} \), we multiply both sides by 4: \[ \cos 2x = \frac{4}{25}. \] To find the general solution for \( 2x \), take the arccosine: \[ 2x = \cos^{-1} \left( \frac{4}{25} \right) + 2k\pi \quad \text{or} \quad 2x = -\cos^{-1} \left( \frac{4}{25} \right) + 2k\pi, \] for any integer \( k \). Dividing both sides by 2 gives: \[ x = \frac{1}{2} \cos^{-1} \left( \frac{4}{25} \right) + k\pi \quad \text{and} \quad x = -\frac{1}{2} \cos^{-1} \left( \frac{4}{25} \right) + k\pi. \]