$$ 1.1 \quad $$ Given $$ \cos \theta=-\frac{1}{3} $$ and $$ 0^{\circ} \leq \theta \leq 180^{\circ} $$ Determine, with the aid of a diagram, the value of the following 1.1.1 $$ \tan \left(180^{\circ}+\theta\right) $$ 1.1.2 $$ 3 \sin \left(\theta-90^{\circ}\right) $$ 1.2 Given: $$ \frac{\sin \left(-210^{\circ}\right)}{\cos \left(300^{\circ}\right)}+\frac{\cos \left(x+90^{\circ}\right)}{\sin \left(360^{\circ}+x\right)} $$ 1.2.1 Simplify the following expression: $$ \quad \frac{\sin \left(-210^{\circ}\right)}{\cos \left(300^{\circ}\right)}+\frac{\cos \left(x+90^{\circ}\right)}{\sin \left(360^{\circ}+x\right)} $$ 1.2.2 For which values of $$ x $$ is the expression in 5.2.1 undefined for $$ -360^{\circ} \leq x \leq 360^{\circ} $$ ? 1.3 Prove that $$ \tan \theta \sqrt{\frac{1}{\sin ^{2} \theta}-1}=1 $$ 1.4 Determine the general solution of the following equation: $$ 2 \sin ^{2} \theta=1+\sin \theta $$ 2.1 Determine, without using a calculator, the value of the following trigonometric expression: $$ \frac{\sin 2 x \cdot \cos (-x)+\cos 2 x \cdot \sin \left(360^{\circ}-x\right)}{\sin \left(180^{\circ}+x\right)} $$ 2.2 Determine the general solution of the following equation: $$ 6 \sin ^{2} x+7 \cos x-3=0 $$ 2.3 Given: $$ x+\frac{1}{x}=3 \cos \mathrm{~A} $$ and $$ x^{2}+\frac{1}{x^{2}}=2 $$ Determine the value of $$ \cos 2 \mathrm{~A} $$ without using a calculator. (5)

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### 1.1 Given $$ \cos \theta = -\frac{1}{3} $$ and $$ 0^{\circ} \leq \theta \leq 180^{\circ} $$

#### 1.1.1 $$ \tan(180^{\circ} + \theta) $$

Using the tangent addition formula:
$$
\tan(180^{\circ} + \theta) = \tan(\theta)
$$
Since $$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$, we need to find $$ \sin(\theta) $$.

Using the Pythagorean identity:
$$
\sin^2(\theta) + \cos^2(\theta) = 1
$$
Substituting $$ \cos(\theta) = -\frac{1}{3} $$:
$$
\sin^2(\theta) + \left(-\frac{1}{3}\right)^2 = 1
$$
$$
\sin^2(\theta) + \frac{1}{9} = 1
$$
$$
\sin^2(\theta) = 1 - \frac{1}{9} = \frac{8}{9}
$$
$$
\sin(\theta) = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \quad (\text{since } \theta \text{ is in the second quadrant})
$$

Now, we can find $$ \tan(\theta) $$:
$$
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{2\sqrt{2}}{3}}{-\frac{1}{3}} = -2\sqrt{2}
$$

Thus,
$$
\tan(180^{\circ} + \theta) = -2\sqrt{2}
$$

#### 1.1.2 $$ 3 \sin(\theta - 90^{\circ}) $$

Using the sine subtraction formula:
$$
\sin(\theta - 90^{\circ}) = -\cos(\theta)
$$
Thus,
$$
3 \sin(\theta - 90^{\circ}) = 3(-\cos(\theta)) = -3\left(-\frac{1}{3}\right) = 1
$$

### 1.2 Given: 
$$
\frac{\sin(-210^{\circ})}{\cos(300^{\circ})} + \frac{\cos(x + 90^{\circ})}{\sin(360^{\circ} + x)}
$$

#### 1.2.1 Simplify the expression

Calculating $$ \sin(-210^{\circ}) $$ and $$ \cos(300^{\circ}) $$:
$$
\sin(-210^{\circ}) = -\sin(210^{\circ}) = -\left(-\frac{1}{2}\right) = \frac{1}{2}
$$
$$
\cos(300^{\circ}) = \frac{1}{2}
$$
Thus,
$$
\frac{\sin(-210^{\circ})}{\cos(300^{\circ})} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1
$$

Now for the second part:
$$
\cos(x + 90^{\circ}) = -\sin(x)
$$
$$
\sin(360^{\circ} + x) = \sin(x)
$$
Thus,
$$
\frac{\cos(x + 90^{\circ})}{\sin(360^{\circ} + x)} = \frac{-\sin(x)}{\sin(x)} = -1 \quad (\text{for } \sin(x) \neq 0)
$$

Combining both parts:
$$
1 - 1 = 0
$$

#### 1.2.2 For which values of $$ x $$ is the expression undefined?

The expression is undefined when $$ \sin(x) = 0 $$:
$$
x = n \cdot 180^{\circ} \quad (n \in \mathbb{Z})
$$
Within the range $$ -360^{\circ} \leq x \leq 360^{\circ} $$, the values are:
$$
-360^{\circ}, -180^{\circ}, 0^{\circ}, 180^{\circ}, 360^{\circ}
$$

### 1.3 Prove that $$ \tan \theta \sqrt{\frac{1}{\sin^2 \theta} - 1} = 1 $$

Starting with the left-hand side:
$$
\tan \theta = \frac{\sin \theta}{\cos \theta}
$$
Thus,
$$
\tan \theta \sqrt{\frac{1}{\sin^2 \theta} - 1} = \frac{\sin \theta}{\cos \theta} \sqrt{\frac{1 - \sin^2 \theta}{\sin^2 \theta}} = \frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} = 1
$$

### 1.4 Determine the general solution of:
$$
2 \sin^2 \theta = 1 + \sin \theta
$$
Rearranging gives:
$$
2 \sin^2 \theta - \sin \theta - 1 = 0
$$
Let $$ y = \sin \theta $$:
$$
2y^2 - y - 1 = 0
$$
Using the quadratic formula:
$$
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}
$$
Thus,
$$
y = 1 \quad \text{or} \quad y = -\frac{1}{2}
$$
So,
$$
\sin \theta = 1 \Rightarrow \theta = 90^{\circ} + 360^{\circ}n
$$
$$
\sin \theta = -\frac{1}{2} \Rightarrow \theta = 210^{\circ} + 360^{\circ}n \quad \text{or} \quad \theta = 330^{\circ} + 360^{\circ}n
$$

### 2.1 Determine the value of:
$$
\frac{\sin 2x \cdot \cos(-x) + \cos 2x \cdot \sin(360^{\circ} - x)}{\sin(180^{\circ} + x)}
$$
Using identities:
\[
\sin(2x) = 2 \sin x \cos x, \quad \cos
1.1.1 $$ \tan(180^{\circ} + \theta) = -2\sqrt{2} $$ 1.1.2 $$ 3 \sin(\theta - 90^{\circ}) = 1 $$ 1.2.1 The expression simplifies to $$ 0 $$. 1.2.2 The expression is undefined when $$ x = 0^{\circ}, 180^{\circ}, 360^{\circ}, -180^{\circ}, -360^{\circ} $$. 1.3 $$ \tan \theta \sqrt{\frac{1}{\sin^2 \theta} - 1} = 1 $$ is proven. 1.4 The general solutions are: - $$ \theta = 90^{\circ} + 360^{\circ}n $$ - $$ \theta = 210^{\circ} + 360^{\circ}n $$ - $$ \theta = 330^{\circ} + 360^{\circ}n $$ 2.1 The value of the expression is $$ 0 $$. 2.2 The general solution is: - $$ x = 60^{\circ} + 360^{\circ}n $$ - $$ x = 300^{\circ} + 360^{\circ}n $$ 2.3 $$ \cos 2A = \frac{1}{2} $$

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