4.2 Determine the general Blution of $$ : $$ $$ 4.2 .1 \quad 2 \cos ^{2} x+7 \sin x=5 $$ $$ 4.2 .2 \quad \cos (x-45)=\sin 15 $$

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Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\cos\left(x-45\right)-\sin\left(15\right)=0$$
- step1: Move the constant to the right side:
  $$\cos\left(x-45\right)=0+\sin\left(15\right)$$
- step2: Add the terms:
  $$\cos\left(x-45\right)=\sin\left(15\right)$$
- step3: Use the inverse trigonometric function:
  $$x-45=\arccos\left(\sin\left(15\right)\right)$$
- step4: Calculate:
  $$\begin{align}&x-45=-\arccos\left(\sin\left(15\right)\right)\&x-45=\arccos\left(\sin\left(15\right)\right)\end{align}$$
- step5: Add the period:
  $$\begin{align}&x-45=-\arccos\left(\sin\left(15\right)\right)+2k\pi ,k \in \mathbb{Z}\&x-45=\arccos\left(\sin\left(15\right)\right)+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step6: Calculate:
  $$\begin{align}&x=-\arccos\left(\sin\left(15\right)\right)+45+2k\pi ,k \in \mathbb{Z}\&x-45=\arccos\left(\sin\left(15\right)\right)+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step7: Calculate:
  $$\begin{align}&x=-\arccos\left(\sin\left(15\right)\right)+45+2k\pi ,k \in \mathbb{Z}\&x=\arccos\left(\sin\left(15\right)\right)+45+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step8: Find the union:
  $$x=\left\{ \begin{array}{l}-\arccos\left(\sin\left(15\right)\right)+45+2k\pi \\arccos\left(\sin\left(15\right)\right)+45+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ 2\cos^{2}x + 7\sin x - 5 = 0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$2\cos^{2}\left(x\right)+7\sin\left(x\right)-5=0$$
- step1: Rewrite the expression:
  $$-3-2\sin^{2}\left(x\right)+7\sin\left(x\right)=0$$
- step2: Factor the expression:
  $$\left(-3+\sin\left(x\right)\right)\left(1-2\sin\left(x\right)\right)=0$$
- step3: Separate into possible cases:
  $$\begin{align}&-3+\sin\left(x\right)=0\&1-2\sin\left(x\right)=0\end{align}$$
- step4: Solve the equation:
  $$\begin{align}&x \notin \mathbb{R}\&x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step5: Find the union:
  $$x=\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Let's solve the two equations step by step.

### 4.2.1: Solve $$ 2 \cos^{2} x + 7 \sin x = 5 $$

1. **Rearranging the equation**:
   $$
   2 \cos^{2} x + 7 \sin x - 5 = 0
   $$
   We can use the identity $$ \cos^{2} x = 1 - \sin^{2} x $$ to express everything in terms of $$ \sin x $$:
   $$
   2(1 - \sin^{2} x) + 7 \sin x - 5 = 0
   $$
   Simplifying this gives:
   $$
   -2 \sin^{2} x + 7 \sin x - 3 = 0
   $$
   Multiplying through by -1:
   $$
   2 \sin^{2} x - 7 \sin x + 3 = 0
   $$

2. **Using the quadratic formula**:
   The solutions for $$ \sin x $$ can be found using the quadratic formula:
   $$
   \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   where $$ a = 2, b = -7, c = 3 $$.

3. **Finding the roots**:
   The roots are:
   $$
   \sin x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}
   $$
   This simplifies to:
   $$
   \sin x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4}
   $$
   Thus, we have:
   $$
   \sin x = 3 \quad \text{(not possible, as } \sin x \text{ must be between -1 and 1)}
   $$
   and
   $$
   \sin x = \frac{1}{2}
   $$

4. **Finding $$ x $$**:
   The solutions for $$ \sin x = \frac{1}{2} $$ are:
   $$
   x = \left\{ \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi \right\}, k \in \mathbb{Z}
   $$

### 4.2.2: Solve $$ \cos(x - 45) = \sin(15) $$

1. **Rearranging the equation**:
   $$
   \cos(x - 45) - \sin(15) = 0
   $$

2. **Using the cosine identity**:
   We can express this as:
   $$
   x - 45 = \pm \arccos(\sin(15)) + 2k\pi
   $$

3. **Finding $$ x $$**:
   Thus, we have:
   $$
   x = -\arccos(\sin(15)) + 45 + 2k\pi
   $$
   and
   $$
   x = \arccos(\sin(15)) + 45 + 2k\pi
   $$

### Final Solutions:
- For $$ 4.2.1 $$:
  $$
  x = \left\{ \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi \right\}, k \in \mathbb{Z}
  $$

- For $$ 4.2.2 $$:
  $$
  x = \left\{ -\arccos(\sin(15)) + 45 + 2k\pi, \arccos(\sin(15)) + 45 + 2k\pi \right\}, k \in \mathbb{Z}
  $$
The solutions are: - For $$ 2 \cos^{2} x + 7 \sin x = 5 $$: $$ x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + 2k\pi \quad \text{for any integer } k $$ - For $$ \cos(x - 45) = \sin(15) $$: $$ x = -\arccos(\sin(15)) + 45 + 2k\pi \quad \text{and} \quad x = \arccos(\sin(15)) + 45 + 2k\pi \quad \text{for any integer } k $$

4.2 Determine the general Blution of \( : \) \( 4.2 .1 \quad 2 \cos ^{2} x+7 \sin x=5 \) \( 4.2 .2 \quad \cos (x-45)=\sin 15 \)

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