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Sure! Let's go through each of the questions one by one.
### Question 15
#### (a) Solve [tex]\(\tan^2 \theta - 3 = 0\)[/tex]
1. Rearrange the equation: [tex]\(\tan^2 \theta = 3\)[/tex].
2. Take the square root of both sides: [tex]\(\tan \theta = \sqrt{3}\)[/tex].
3. Using the known angle values, [tex]\(\tan \theta = \sqrt{3}\)[/tex] corresponds to [tex]\(\theta = 60^\circ\)[/tex].
So, [tex]\(\theta = 60^\circ\)[/tex].
#### (d) Solve [tex]\((2 \sin \theta - 1)(\sin \theta - 2) = 0\)[/tex]
1. Set each factor to zero:
- [tex]\(2 \sin \theta - 1 = 0 \implies \sin \theta = \frac{1}{2}\)[/tex]
- [tex]\(\sin \theta - 2 = 0 \implies \sin \theta = 2\)[/tex]
2. [tex]\(\sin \theta = \frac{1}{2}\)[/tex] corresponds to [tex]\(\theta = 30^\circ\)[/tex].
- Note: Since [tex]\(\sin \theta\)[/tex] cannot be greater than 1, [tex]\(\sin \theta = 2\)[/tex] is not a valid solution.
So, [tex]\(\theta = 30^\circ\)[/tex].
#### (c) Solve [tex]\(2 \cos^2 \theta - 1 = 0\)[/tex]
1. Rearrange the equation: [tex]\(2 \cos^2 \theta = 1 \implies \cos^2 \theta = \frac{1}{2}\)[/tex].
2. Take the square root of both sides: [tex]\(\cos \theta = \frac{\sqrt{2}}{2}\)[/tex].
[tex]\(\cos \theta = \frac{\sqrt{2}}{2}\)[/tex] corresponds to [tex]\(\theta = 45^\circ\)[/tex].
So, [tex]\(\theta = 45^\circ\)[/tex].
#### (e) Solve [tex]\(2 \cos^2 \theta + \cos \theta - 1 = 0\)[/tex]
This is a quadratic equation in [tex]\(\cos \theta\)[/tex]. Let [tex]\(x = \cos \theta\)[/tex]:
1. The quadratic equation becomes: [tex]\(2x^2 + x - 1 = 0\)[/tex].
2. Factoring the quadratic equation, we get: [tex]\((2x - 1)(x + 1) = 0\)[/tex].
3. Solve for [tex]\(x\)[/tex]:
- [tex]\(2x - 1 = 0 \implies x = \frac{1}{2}\)[/tex]
- [tex]\(x + 1 = 0 \implies x = -1\)[/tex]
For [tex]\(0^\circ \leq \theta \leq 90^\circ\)[/tex]:
- [tex]\(\cos \theta = \frac{1}{2}\)[/tex] corresponds to [tex]\(\theta = 60^\circ\)[/tex].
- [tex]\(\cos \theta = -1\)[/tex] is not valid as [tex]\(\theta\)[/tex] must be within [tex]\(0^\circ\)[/tex] to [tex]\(90^\circ\)[/tex].
So, [tex]\(\theta = 60^\circ\)[/tex].
### Question 16
Given [tex]\(2 \sin^2 \theta - 1 = 0\)[/tex]:
1. Rearrange the equation: [tex]\(\sin^2 \theta = \frac{1}{2}\)[/tex].
2. Take the square root: [tex]\(\sin \theta = \frac{\sqrt{2}}{2}\)[/tex].
When [tex]\(\sin \theta = \frac{\sqrt{2}}{2}\)[/tex], [tex]\(\theta = 45^\circ\)[/tex].
#### (a) Find [tex]\(\cos \theta + \cos 2\theta\)[/tex]
- [tex]\(\cos 45^\circ = \frac{\sqrt{2}}{2}\)[/tex].
- [tex]\(\cos 2\theta = \cos 90^\circ = 0\)[/tex].
So, [tex]\(\cos \theta + \cos 2\theta = \frac{\sqrt{2}}{2} + 0 = \frac{\sqrt{2}}{2}\)[/tex].
#### (b) Find [tex]\(\sin \theta \cdot \sin 2\theta\)[/tex]
- [tex]\(\sin 45^\circ = \frac{\sqrt{2}}{2}\)[/tex].
- [tex]\(\sin 2\theta = \sin 90^\circ = 1\)[/tex].
So, [tex]\(\sin \theta \cdot \sin 2\theta = \frac{\sqrt{2}}{2} \cdot 1 = \frac{\sqrt{2}}{2}\)[/tex].
### Question 17
Given [tex]\(\theta\)[/tex] is an acute angle and [tex]\(\cos \theta = \sin \theta\)[/tex]:
1. [tex]\(\cos \theta = \sin \theta \implies \theta = 45^\circ\)[/tex].
#### (a) Find [tex]\(2 \tan^2 \theta + \cos^2 \theta - 1\)[/tex]
- [tex]\(\theta = 45^\circ\)[/tex].
- [tex]\(\tan 45^\circ = 1\)[/tex], so [tex]\(\tan^2 45^\circ = 1\)[/tex].
- [tex]\(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\)[/tex], so [tex]\(\cos^2 45^\circ = \frac{1}{2}\)[/tex].
Thus,
[tex]\[ 2 \tan^2 \theta + \cos^2 \theta - 1 = 2 \cdot 1 + \frac{1}{2} - 1 = 2 + \frac{1}{2} - 1 = 1 + \frac{1}{2} = \frac{3}{2}. \][/tex]
#### (b) Find [tex]\(2 \cos^2 \theta - 1\)[/tex]
- [tex]\(\cos 45^\circ = \frac{\sqrt{2}}{2}\)[/tex], so [tex]\(\cos^2 45^\circ = \frac{1}{2}\)[/tex].
Thus,
[tex]\[ 2 \cos^2 \theta - 1 = 2 \cdot \frac{1}{2} - 1 = 1 - 1 = 0. \][/tex]
With these detailed steps, we have solved all the problems.
### Question 15
#### (a) Solve [tex]\(\tan^2 \theta - 3 = 0\)[/tex]
1. Rearrange the equation: [tex]\(\tan^2 \theta = 3\)[/tex].
2. Take the square root of both sides: [tex]\(\tan \theta = \sqrt{3}\)[/tex].
3. Using the known angle values, [tex]\(\tan \theta = \sqrt{3}\)[/tex] corresponds to [tex]\(\theta = 60^\circ\)[/tex].
So, [tex]\(\theta = 60^\circ\)[/tex].
#### (d) Solve [tex]\((2 \sin \theta - 1)(\sin \theta - 2) = 0\)[/tex]
1. Set each factor to zero:
- [tex]\(2 \sin \theta - 1 = 0 \implies \sin \theta = \frac{1}{2}\)[/tex]
- [tex]\(\sin \theta - 2 = 0 \implies \sin \theta = 2\)[/tex]
2. [tex]\(\sin \theta = \frac{1}{2}\)[/tex] corresponds to [tex]\(\theta = 30^\circ\)[/tex].
- Note: Since [tex]\(\sin \theta\)[/tex] cannot be greater than 1, [tex]\(\sin \theta = 2\)[/tex] is not a valid solution.
So, [tex]\(\theta = 30^\circ\)[/tex].
#### (c) Solve [tex]\(2 \cos^2 \theta - 1 = 0\)[/tex]
1. Rearrange the equation: [tex]\(2 \cos^2 \theta = 1 \implies \cos^2 \theta = \frac{1}{2}\)[/tex].
2. Take the square root of both sides: [tex]\(\cos \theta = \frac{\sqrt{2}}{2}\)[/tex].
[tex]\(\cos \theta = \frac{\sqrt{2}}{2}\)[/tex] corresponds to [tex]\(\theta = 45^\circ\)[/tex].
So, [tex]\(\theta = 45^\circ\)[/tex].
#### (e) Solve [tex]\(2 \cos^2 \theta + \cos \theta - 1 = 0\)[/tex]
This is a quadratic equation in [tex]\(\cos \theta\)[/tex]. Let [tex]\(x = \cos \theta\)[/tex]:
1. The quadratic equation becomes: [tex]\(2x^2 + x - 1 = 0\)[/tex].
2. Factoring the quadratic equation, we get: [tex]\((2x - 1)(x + 1) = 0\)[/tex].
3. Solve for [tex]\(x\)[/tex]:
- [tex]\(2x - 1 = 0 \implies x = \frac{1}{2}\)[/tex]
- [tex]\(x + 1 = 0 \implies x = -1\)[/tex]
For [tex]\(0^\circ \leq \theta \leq 90^\circ\)[/tex]:
- [tex]\(\cos \theta = \frac{1}{2}\)[/tex] corresponds to [tex]\(\theta = 60^\circ\)[/tex].
- [tex]\(\cos \theta = -1\)[/tex] is not valid as [tex]\(\theta\)[/tex] must be within [tex]\(0^\circ\)[/tex] to [tex]\(90^\circ\)[/tex].
So, [tex]\(\theta = 60^\circ\)[/tex].
### Question 16
Given [tex]\(2 \sin^2 \theta - 1 = 0\)[/tex]:
1. Rearrange the equation: [tex]\(\sin^2 \theta = \frac{1}{2}\)[/tex].
2. Take the square root: [tex]\(\sin \theta = \frac{\sqrt{2}}{2}\)[/tex].
When [tex]\(\sin \theta = \frac{\sqrt{2}}{2}\)[/tex], [tex]\(\theta = 45^\circ\)[/tex].
#### (a) Find [tex]\(\cos \theta + \cos 2\theta\)[/tex]
- [tex]\(\cos 45^\circ = \frac{\sqrt{2}}{2}\)[/tex].
- [tex]\(\cos 2\theta = \cos 90^\circ = 0\)[/tex].
So, [tex]\(\cos \theta + \cos 2\theta = \frac{\sqrt{2}}{2} + 0 = \frac{\sqrt{2}}{2}\)[/tex].
#### (b) Find [tex]\(\sin \theta \cdot \sin 2\theta\)[/tex]
- [tex]\(\sin 45^\circ = \frac{\sqrt{2}}{2}\)[/tex].
- [tex]\(\sin 2\theta = \sin 90^\circ = 1\)[/tex].
So, [tex]\(\sin \theta \cdot \sin 2\theta = \frac{\sqrt{2}}{2} \cdot 1 = \frac{\sqrt{2}}{2}\)[/tex].
### Question 17
Given [tex]\(\theta\)[/tex] is an acute angle and [tex]\(\cos \theta = \sin \theta\)[/tex]:
1. [tex]\(\cos \theta = \sin \theta \implies \theta = 45^\circ\)[/tex].
#### (a) Find [tex]\(2 \tan^2 \theta + \cos^2 \theta - 1\)[/tex]
- [tex]\(\theta = 45^\circ\)[/tex].
- [tex]\(\tan 45^\circ = 1\)[/tex], so [tex]\(\tan^2 45^\circ = 1\)[/tex].
- [tex]\(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\)[/tex], so [tex]\(\cos^2 45^\circ = \frac{1}{2}\)[/tex].
Thus,
[tex]\[ 2 \tan^2 \theta + \cos^2 \theta - 1 = 2 \cdot 1 + \frac{1}{2} - 1 = 2 + \frac{1}{2} - 1 = 1 + \frac{1}{2} = \frac{3}{2}. \][/tex]
#### (b) Find [tex]\(2 \cos^2 \theta - 1\)[/tex]
- [tex]\(\cos 45^\circ = \frac{\sqrt{2}}{2}\)[/tex], so [tex]\(\cos^2 45^\circ = \frac{1}{2}\)[/tex].
Thus,
[tex]\[ 2 \cos^2 \theta - 1 = 2 \cdot \frac{1}{2} - 1 = 1 - 1 = 0. \][/tex]
With these detailed steps, we have solved all the problems.
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