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Sagot :
Certainly! Let's solve each part of the given problem step-by-step.
### Part (a): Solve the equation [tex]\(\sin \theta + \cos \theta + 1 = 0\)[/tex]
First, let's manipulate the given equation:
[tex]\[ \sin \theta + \cos \theta + 1 = 0 \][/tex]
Rearranging terms, we get:
[tex]\[ \sin \theta + \cos \theta = -1 \][/tex]
We know from trigonometric identities that:
[tex]\[ \sin \theta + \cos \theta = \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \][/tex]
However, let's look at the specific solution provided:
Given that [tex]\(\theta = -\frac{\pi}{2}\)[/tex] is a solution. We want to provide the general solution for [tex]\(\theta\)[/tex]. The sine and cosine functions are periodic with period [tex]\(2\pi\)[/tex], but we must account for their specific combined behavior in this sum.
The general solution for [tex]\(\theta\)[/tex] is:
[tex]\[ \theta = -\frac{\pi}{2} + 2k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
Now, let's move on to the next part.
### Part (b): Solve the equation [tex]\(\tan^3 \theta - 3 \tan \theta = 0\)[/tex]
We start with the given equation:
[tex]\[ \tan^3 \theta - 3 \tan \theta = 0 \][/tex]
Factorizing, we get:
[tex]\[ \tan \theta (\tan^2 \theta - 3) = 0 \][/tex]
This results in two separate equations:
1. [tex]\(\tan \theta = 0\)[/tex]
2. [tex]\(\tan^2 \theta = 3\)[/tex]
For [tex]\(\tan \theta = 0\)[/tex]:
[tex]\[ \theta = k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
For [tex]\(\tan^2 \theta = 3\)[/tex]:
[tex]\[ \tan \theta = \pm \sqrt{3} \][/tex]
Solving for [tex]\(\theta\)[/tex], we have:
[tex]\[ \theta = \pm \frac{\pi}{3} + k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
Thus, combining all possible solutions, we have the general solution for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = k\pi, \quad \theta = \frac{\pi}{3} + k\pi, \quad \text{or} \quad \theta = -\frac{\pi}{3} + k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
### Summary of General Solutions:
- For [tex]\(\sin \theta + \cos \theta + 1 = 0\)[/tex]:
[tex]\[ \theta = -\frac{\pi}{2} + 2k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
- For [tex]\(\tan^3 \theta - 3 \tan \theta = 0\)[/tex]:
[tex]\[ \theta = k\pi, \quad \theta = \frac{\pi}{3} + k\pi, \quad \text{or} \quad \theta = -\frac{\pi}{3} + k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
### Part (a): Solve the equation [tex]\(\sin \theta + \cos \theta + 1 = 0\)[/tex]
First, let's manipulate the given equation:
[tex]\[ \sin \theta + \cos \theta + 1 = 0 \][/tex]
Rearranging terms, we get:
[tex]\[ \sin \theta + \cos \theta = -1 \][/tex]
We know from trigonometric identities that:
[tex]\[ \sin \theta + \cos \theta = \sqrt{2} \sin \left( \theta + \frac{\pi}{4} \right) \][/tex]
However, let's look at the specific solution provided:
Given that [tex]\(\theta = -\frac{\pi}{2}\)[/tex] is a solution. We want to provide the general solution for [tex]\(\theta\)[/tex]. The sine and cosine functions are periodic with period [tex]\(2\pi\)[/tex], but we must account for their specific combined behavior in this sum.
The general solution for [tex]\(\theta\)[/tex] is:
[tex]\[ \theta = -\frac{\pi}{2} + 2k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
Now, let's move on to the next part.
### Part (b): Solve the equation [tex]\(\tan^3 \theta - 3 \tan \theta = 0\)[/tex]
We start with the given equation:
[tex]\[ \tan^3 \theta - 3 \tan \theta = 0 \][/tex]
Factorizing, we get:
[tex]\[ \tan \theta (\tan^2 \theta - 3) = 0 \][/tex]
This results in two separate equations:
1. [tex]\(\tan \theta = 0\)[/tex]
2. [tex]\(\tan^2 \theta = 3\)[/tex]
For [tex]\(\tan \theta = 0\)[/tex]:
[tex]\[ \theta = k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
For [tex]\(\tan^2 \theta = 3\)[/tex]:
[tex]\[ \tan \theta = \pm \sqrt{3} \][/tex]
Solving for [tex]\(\theta\)[/tex], we have:
[tex]\[ \theta = \pm \frac{\pi}{3} + k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
Thus, combining all possible solutions, we have the general solution for [tex]\(\theta\)[/tex]:
[tex]\[ \theta = k\pi, \quad \theta = \frac{\pi}{3} + k\pi, \quad \text{or} \quad \theta = -\frac{\pi}{3} + k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
### Summary of General Solutions:
- For [tex]\(\sin \theta + \cos \theta + 1 = 0\)[/tex]:
[tex]\[ \theta = -\frac{\pi}{2} + 2k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
- For [tex]\(\tan^3 \theta - 3 \tan \theta = 0\)[/tex]:
[tex]\[ \theta = k\pi, \quad \theta = \frac{\pi}{3} + k\pi, \quad \text{or} \quad \theta = -\frac{\pi}{3} + k\pi \quad \text{for} \quad k \in \mathbb{Z} \][/tex]
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