QUEDIIUN 12 \( \begin{array}{l}12.1 \text { Prove that, if } \cos (\alpha-x) \neq 0 \\ \frac{\sin \left(x+450^{\circ}-\alpha\right)}{\cos (\alpha-x)}\end{array}=1 \) (3) \( \begin{array}{l}\text { Determine the general solution of } \cos 2 x=1-3 \cos x\end{array} \)

**Step 12.1** We need to prove that if \[ \cos(\alpha-x) \neq 0, \] then \[ \frac{\sin\left(x+450^\circ-\alpha\right)}{\cos(\alpha-x)} = 1. \] 1. Notice that \[ 450^\circ = 360^\circ + 90^\circ. \] Therefore, \[ x + 450^\circ - \alpha = x + 360^\circ + 90^\circ - \alpha. \] 2. Since adding \(360^\circ\) does not change the value of a trigonometric function, we have \[ \sin\left(x + 450^\circ - \alpha\right) = \sin\left(x + 90^\circ - \alpha\right). \] 3. Rewrite the angle as \[ x + 90^\circ - \alpha = (x-\alpha) + 90^\circ. \] 4. Use the well-known trigonometric identity \[ \sin\left(90^\circ + \theta\right) = \cos \theta. \] Here, letting \(\theta = x-\alpha\) gives \[ \sin\left((x-\alpha)+90^\circ\right) = \cos(x-\alpha). \] 5. Thus, we have \[ \sin\left(x+450^\circ-\alpha\right) = \cos(x-\alpha). \] 6. Notice that the cosine function is even, that is, \[ \cos(x-\alpha) = \cos(\alpha-x). \] 7. Therefore, the original fraction simplifies to \[ \frac{\sin\left(x+450^\circ-\alpha\right)}{\cos(\alpha-x)} = \frac{\cos(x-\alpha)}{\cos(\alpha-x)} = \frac{\cos(\alpha-x)}{\cos(\alpha-x)} = 1, \] as long as \(\cos(\alpha-x) \neq 0\). --- **Step (3): Determine the General Solution of \(\cos2x = 1-3\cos x\)** 1. Recall the double-angle formula for cosine: \[ \cos2x = 2\cos^2x - 1. \] Substitute this into the given equation: \[ 2\cos^2x - 1 = 1 - 3\cos x. \] 2. Rearrange the equation to bring all terms to one side: \[ 2\cos^2 x - 1 - 1 + 3\cos x = 0, \] which simplifies to \[ 2\cos^2 x + 3\cos x - 2 = 0. \] 3. Let \(u = \cos x\). Then the equation becomes: \[ 2u^2 + 3u - 2 = 0. \] 4. Solve the quadratic equation using the quadratic formula: \[ u = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4}. \] Thus, \[ u = \frac{-3 \pm 5}{4}. \] 5. This gives two possible solutions: - \(u = \frac{-3+5}{4} = \frac{2}{4} = \frac{1}{2}\), - \(u = \frac{-3-5}{4} = \frac{-8}{4} = -2\). 6. Since the range of the cosine function is \([-1,1]\), the solution \(u = -2\) is extraneous. Therefore, we only have \[ \cos x = \frac{1}{2}. \] 7. The general solution for \(\cos x = \frac{1}{2}\) in degrees is: \[ x = 60^\circ + 360^\circ k \quad \text{or} \quad x = 300^\circ + 360^\circ k, \quad k \in \mathbb{Z}, \] or in radians: \[ x = \frac{\pi}{3} + 2\pi k \quad \text{or} \quad x = \frac{5\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}. \]