Classwork Solving trigonome for $$ heta $$ where $$ heta \in $$ 1. $$ \cos \left( heta-30^{\circ} ight)=-0,5 $$ 2. $$ 2 \sin heta \cdot \cos heta-\cos heta= $$ 3. $$ 2 \cos ^{2} heta-7 \cos heta-4 $$

Question

Classwork Solving trigonome for $$ 	heta $$ where $$ 	heta \in $$ 1. $$ \cos \left(	heta-30^{\circ}ight)=-0,5 $$ 2. $$ 2 \sin 	heta \cdot \cos 	heta-\cos 	heta= $$ 3. $$ 2 \cos ^{2} 	heta-7 \cos 	heta-4 $$

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Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$\cos\left(\theta -30^{\circ}\right)=-0.5$$
- step1: Use the periodicity identities:
  $$\cos\left(\theta +330^{\circ}\right)=-0.5$$
- step2: Use the inverse trigonometric function:
  $$\theta +330^{\circ}=\arccos\left(-0.5\right)$$
- step3: Calculate:
  $$\begin{align}&\theta +330^{\circ}=\frac{2\pi }{3}\&\theta +330^{\circ}=\frac{4\pi }{3}\end{align}$$
- step4: Add the period:
  $$\begin{align}&\theta +330^{\circ}=\frac{2\pi }{3}+2k\pi ,k \in \mathbb{Z}\&\theta +330^{\circ}=\frac{4\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step5: Calculate:
  $$\begin{align}&\theta =\frac{5\pi }{6}+2k\pi ,k \in \mathbb{Z}\&\theta +330^{\circ}=\frac{4\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step6: Calculate:
  $$\begin{align}&\theta =\frac{5\pi }{6}+2k\pi ,k \in \mathbb{Z}\&\theta =\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\end{align}$$
- step7: Find the union:
  $$\theta =\left\{ \begin{array}{l}\frac{5\pi }{6}+2k\pi \\frac{3\pi }{2}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ 2 \sin(\theta) \cdot \cos(\theta) - \cos(\theta) = 0 $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$2\sin\left(\theta \right)\cos\left(\theta \right)-\cos\left(\theta \right)=0$$
- step1: Factor the expression:
  $$\cos\left(\theta \right)\left(2\sin\left(\theta \right)-1\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&\cos\left(\theta \right)=0\&2\sin\left(\theta \right)-1=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&\theta =\frac{\pi }{2}+k\pi ,k \in \mathbb{Z}\&\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step4: Find the union:
  $$\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\frac{\pi }{2}+k\pi \\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ 2 \cos^{2}(\theta) - 7 \cos(\theta) - 4 = 0 $$.
Solve the equation by following steps:
- step0: Solve for $$\theta$$:
  $$2\cos^{2}\left(\theta \right)-7\cos\left(\theta \right)-4=0$$
- step1: Factor the expression:
  $$\left(\cos\left(\theta \right)-4\right)\left(2\cos\left(\theta \right)+1\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&\cos\left(\theta \right)-4=0\&2\cos\left(\theta \right)+1=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&\theta  \notin \mathbb{R}\&\theta =\left\{ \begin{array}{l}\frac{2\pi }{3}+2k\pi \\frac{4\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}$$
- step4: Find the union:
  $$\theta =\left\{ \begin{array}{l}\frac{2\pi }{3}+2k\pi \\frac{4\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}$$
Let's solve each of the trigonometric equations step by step.

### 1. Solve $$ \cos(\theta - 30^{\circ}) = -0.5 $$

To solve this equation, we recognize that $$ \cos(\theta) = -0.5 $$ corresponds to specific angles. The general solutions for $$ \cos(x) = -0.5 $$ are:

$$
x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}
$$

Substituting back for $$ \theta - 30^{\circ} $$:

$$
\theta - 30^{\circ} = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad \theta - 30^{\circ} = \frac{4\pi}{3} + 2k\pi
$$

Thus, we have:

$$
\theta = \frac{2\pi}{3} + 30^{\circ} + 2k\pi \quad \text{and} \quad \theta = \frac{4\pi}{3} + 30^{\circ} + 2k\pi
$$

The solutions are:

$$
\theta = \left\{ \frac{5\pi}{6} + 2k\pi, \frac{3\pi}{2} + 2k\pi \right\}, \quad k \in \mathbb{Z}
$$

### 2. Solve $$ 2 \sin \theta \cdot \cos \theta - \cos \theta = 0 $$

Factoring out $$ \cos \theta $$:

$$
\cos \theta (2 \sin \theta - 1) = 0
$$

This gives us two cases to solve:

1. $$ \cos \theta = 0 $$
2. $$ 2 \sin \theta - 1 = 0 $$

For $$ \cos \theta = 0 $$:

$$
\theta = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}
$$

For $$ 2 \sin \theta - 1 = 0 $$:

$$
\sin \theta = \frac{1}{2}
$$

The general solutions for $$ \sin \theta = \frac{1}{2} $$ are:

$$
\theta = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad \theta = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}
$$

Combining both cases, we have:

$$
\theta = \left\{ \frac{\pi}{6} + 2k\pi, \frac{\pi}{2} + k\pi, \frac{5\pi}{6} + 2k\pi \right\}, \quad k \in \mathbb{Z}
$$

### 3. Solve $$ 2 \cos^{2} \theta - 7 \cos \theta - 4 = 0 $$

This is a quadratic equation in terms of $$ \cos \theta $$. Let $$ x = \cos \theta $$:

$$
2x^2 - 7x - 4 = 0
$$

Using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$:

$$
x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2}
$$

Calculating the discriminant:

$$
49 + 32 = 81
$$

Thus, we have:

$$
x = \frac{7 \pm 9}{4}
$$

Calculating the two possible values for $$ x $$:

1. $$ x = \frac{16}{4} = 4 $$ (not valid since $$ \cos \theta $$ must be in $$[-1, 1]$$)
2. $$ x = \frac{-2}{4} = -\frac{1}{2} $$

Now, solving $$ \cos \theta = -\frac{1}{2} $$:

The general solutions for $$ \cos \theta = -\frac{1}{2} $$ are:

$$
\theta = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}
$$

### Summary of Solutions

1. $$ \theta = \left\{ \frac{5\pi}{6} + 2k\pi, \frac{3\pi}{2} + 2k\pi \right\}, \quad k \in \mathbb{Z} $$
2. $$ \theta = \left\{ \frac{\pi}{6} + 2k\pi, \frac{\pi}{2} + k\pi, \frac{5\pi}{6} + 2k\pi \right\}, \quad k \in \mathbb{Z} $$
3. $$ \theta = \left\{ \frac{2\pi}{3} + 2k\pi, \frac{4\pi}{3} + 2k\pi \right\}, \quad k \in \mathbb{Z} $$
1. $$ \theta = \frac{5\pi}{6} + 2k\pi $$ and $$ \theta = \frac{3\pi}{2} + 2k\pi $$, where $$ k $$ is an integer. 2. $$ \theta = \frac{\pi}{6} + 2k\pi $$, $$ \theta = \frac{\pi}{2} + k\pi $$, and $$ \theta = \frac{5\pi}{6} + 2k\pi $$, where $$ k $$ is an integer. 3. $$ \theta = \frac{2\pi}{3} + 2k\pi $$ and $$ \theta = \frac{4\pi}{3} + 2k\pi $$, where $$ k $$ is an integer.

Classwork Solving trigonome for \( \theta \) where \( \theta \in \) 1. \( \cos \left(\theta-30^{\circ}\right)=-0,5 \) 2. \( 2 \sin \theta \cdot \cos \theta-\cos \theta= \) 3. \( 2 \cos ^{2} \theta-7 \cos \theta-4 \)

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