IDNLearn.com offers a user-friendly platform for finding and sharing answers. Discover comprehensive answers to your questions from our community of knowledgeable experts.
Sagot :
To find the general solution to the trigonometric equation [tex]\(2 \cos^2 \theta - \cos \theta = 0\)[/tex], let's go through the steps in detail:
1. Factor the equation:
The given equation is:
[tex]\[ 2 \cos^2 \theta - \cos \theta = 0 \][/tex]
Factor out [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos \theta (2 \cos \theta - 1) = 0 \][/tex]
2. Set each factor to zero and solve:
For [tex]\(\cos \theta = 0\)[/tex]:
[tex]\[ \cos \theta = 0 \][/tex]
The angles [tex]\(\theta\)[/tex] where [tex]\(\cos \theta = 0\)[/tex] can be found using the unit circle. These angles are:
[tex]\[ \theta = 90^\circ + 360^\circ k \quad \text{or} \quad \theta = 270^\circ + 360^\circ k \quad \text{where } k \in \mathbb{Z} \][/tex]
For [tex]\(2 \cos \theta - 1 = 0\)[/tex]:
[tex]\[ 2 \cos \theta - 1 = 0 \quad \Rightarrow \quad 2 \cos \theta = 1 \quad \Rightarrow \quad \cos \theta = \frac{1}{2} \][/tex]
The angles [tex]\(\theta\)[/tex] where [tex]\(\cos \theta = \frac{1}{2}\)[/tex] can be found using the unit circle. These angles are:
[tex]\[ \theta = 60^\circ + 360^\circ k \quad \text{or} \quad \theta = 300^\circ + 360^\circ k \quad \text{where } k \in \mathbb{Z} \][/tex]
3. Combine all general solutions:
Therefore, the general solutions to the equation [tex]\(2 \cos^2 \theta - \cos \theta = 0\)[/tex] are:
[tex]\[ \theta = 90^\circ + 360^\circ k \quad \text{or} \quad \theta = 270^\circ + 360^\circ k \quad \text{or} \quad \theta = 60^\circ + 360^\circ k \quad \text{or} \quad \theta = 300^\circ + 360^\circ k \][/tex]
where [tex]\(k \in \mathbb{Z}\)[/tex].
4. Match the solution with the choices provided:
Comparing with the given choices:
None of the choices exactly match the derived general solutions. Let's identify the proper captures:
a. C. [tex]$\theta=90^{\circ}+360 k$[/tex] or [tex]$\theta=270^{\circ}+360 k$[/tex] and [tex]$\theta=60^{\circ}+360 k$[/tex] or [tex]$\theta=300^{\circ}+360 k$[/tex] where [tex]$k \in Z$[/tex]
Does precisely match our combination of solutions.
Thus, the correct answer is:
a. [tex]\( \theta=90^\circ + 360k \text{ or } \theta=270^\circ + 360k \text{ and } \theta=60^\circ + 360k \text{ or } \theta=300^\circ + 360k \text{ where } k \in \mathbb{Z} \)[/tex]
1. Factor the equation:
The given equation is:
[tex]\[ 2 \cos^2 \theta - \cos \theta = 0 \][/tex]
Factor out [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos \theta (2 \cos \theta - 1) = 0 \][/tex]
2. Set each factor to zero and solve:
For [tex]\(\cos \theta = 0\)[/tex]:
[tex]\[ \cos \theta = 0 \][/tex]
The angles [tex]\(\theta\)[/tex] where [tex]\(\cos \theta = 0\)[/tex] can be found using the unit circle. These angles are:
[tex]\[ \theta = 90^\circ + 360^\circ k \quad \text{or} \quad \theta = 270^\circ + 360^\circ k \quad \text{where } k \in \mathbb{Z} \][/tex]
For [tex]\(2 \cos \theta - 1 = 0\)[/tex]:
[tex]\[ 2 \cos \theta - 1 = 0 \quad \Rightarrow \quad 2 \cos \theta = 1 \quad \Rightarrow \quad \cos \theta = \frac{1}{2} \][/tex]
The angles [tex]\(\theta\)[/tex] where [tex]\(\cos \theta = \frac{1}{2}\)[/tex] can be found using the unit circle. These angles are:
[tex]\[ \theta = 60^\circ + 360^\circ k \quad \text{or} \quad \theta = 300^\circ + 360^\circ k \quad \text{where } k \in \mathbb{Z} \][/tex]
3. Combine all general solutions:
Therefore, the general solutions to the equation [tex]\(2 \cos^2 \theta - \cos \theta = 0\)[/tex] are:
[tex]\[ \theta = 90^\circ + 360^\circ k \quad \text{or} \quad \theta = 270^\circ + 360^\circ k \quad \text{or} \quad \theta = 60^\circ + 360^\circ k \quad \text{or} \quad \theta = 300^\circ + 360^\circ k \][/tex]
where [tex]\(k \in \mathbb{Z}\)[/tex].
4. Match the solution with the choices provided:
Comparing with the given choices:
None of the choices exactly match the derived general solutions. Let's identify the proper captures:
a. C. [tex]$\theta=90^{\circ}+360 k$[/tex] or [tex]$\theta=270^{\circ}+360 k$[/tex] and [tex]$\theta=60^{\circ}+360 k$[/tex] or [tex]$\theta=300^{\circ}+360 k$[/tex] where [tex]$k \in Z$[/tex]
Does precisely match our combination of solutions.
Thus, the correct answer is:
a. [tex]\( \theta=90^\circ + 360k \text{ or } \theta=270^\circ + 360k \text{ and } \theta=60^\circ + 360k \text{ or } \theta=300^\circ + 360k \text{ where } k \in \mathbb{Z} \)[/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.