Get comprehensive solutions to your problems with IDNLearn.com. Get accurate and detailed answers to your questions from our dedicated community members who are always ready to help.
Sagot :
Certainly! Let's break this problem down step by step.
### Part (a)
Given the equation:
[tex]\[ 5 \sin \theta = 2 \cos \theta \][/tex]
We need to find [tex]\(\tan \theta\)[/tex].
1. Recall that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
2. From the given equation:
[tex]\[ 5 \sin \theta = 2 \cos \theta \][/tex]
3. We can divide both sides by [tex]\(\cos \theta\)[/tex] (assuming [tex]\(\cos \theta \neq 0\)[/tex]):
[tex]\[ 5 \frac{\sin \theta}{\cos \theta} = 2 \][/tex]
[tex]\[ 5 \tan \theta = 2 \][/tex]
4. Solving for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{2}{5} \][/tex]
So the value of [tex]\(\tan \theta\)[/tex] is:
[tex]\[ \boxed{0.4} \][/tex]
### Part (b)
Given the equation:
[tex]\[ 5 \sin 2x = 2 \cos 2x \][/tex]
We need to solve for [tex]\(x\)[/tex] in the range [tex]\(0 \leq x < 360^\circ\)[/tex].
1. First, recall that [tex]\(\tan 2x = \frac{\sin 2x}{\cos 2x}\)[/tex]. Therefore, we can rewrite the given equation as:
[tex]\[ 5 \sin 2x = 2 \cos 2x \][/tex]
2. Dividing both sides by [tex]\(\cos 2x\)[/tex] (assuming [tex]\(\cos 2x \neq 0\)[/tex]):
[tex]\[ 5 \frac{\sin 2x}{\cos 2x} = 2 \][/tex]
[tex]\[ 5 \tan 2x = 2 \][/tex]
3. Solving for [tex]\(\tan 2x\)[/tex]:
[tex]\[ \tan 2x = \frac{2}{5} \][/tex]
4. To find the solutions for [tex]\(x\)[/tex], consider the general solution for [tex]\(\tan \phi = k\)[/tex]:
[tex]\[ \phi = \tan^{-1}(k) + n \cdot 180^\circ, \; n \in \mathbb{Z} \][/tex]
Here, [tex]\(\phi = 2x\)[/tex] and [tex]\(k = \frac{2}{5}\)[/tex].
5. Thus, the general solution for [tex]\(2x\)[/tex] is:
[tex]\[ 2x = \tan^{-1}\left(\frac{2}{5}\right) + k \cdot 180^\circ \][/tex]
6. Dividing by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{1}{2} \tan^{-1}\left(\frac{2}{5}\right) + n \cdot 90^\circ \][/tex]
7. Calculate the principal value of [tex]\(\tan^{-1}\left(\frac{2}{5}\right)\)[/tex]. It is approximately [tex]\(21.8^\circ\)[/tex]. Therefore:
[tex]\[ x = \frac{21.8^\circ}{2} + n \cdot 90^\circ \][/tex]
[tex]\[ x = 10.9^\circ + n \cdot 90^\circ \][/tex]
8. Enumerate the solutions within the given range [tex]\(0 \leq x < 360^\circ\)[/tex]:
- For [tex]\(n = 0\)[/tex], [tex]\( x = 10.9^\circ \)[/tex]
- For [tex]\(n = 1\)[/tex], [tex]\( x = 10.9^\circ + 90^\circ = 100.9^\circ \)[/tex]
- For [tex]\(n = 2\)[/tex], [tex]\( x = 10.9^\circ + 180^\circ = 190.9^\circ \)[/tex]
- For [tex]\(n = 3\)[/tex], [tex]\( x = 10.9^\circ + 270^\circ = 280.9^\circ \)[/tex]
Thus, the solutions for [tex]\(x\)[/tex] in the range [tex]\(0 \leq x < 360^\circ\)[/tex] are:
[tex]\[ \boxed{10.9^\circ, 100.9^\circ, 190.9^\circ, 280.9^\circ} \][/tex]
These are all the steps to solve the given problem correctly.
### Part (a)
Given the equation:
[tex]\[ 5 \sin \theta = 2 \cos \theta \][/tex]
We need to find [tex]\(\tan \theta\)[/tex].
1. Recall that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
2. From the given equation:
[tex]\[ 5 \sin \theta = 2 \cos \theta \][/tex]
3. We can divide both sides by [tex]\(\cos \theta\)[/tex] (assuming [tex]\(\cos \theta \neq 0\)[/tex]):
[tex]\[ 5 \frac{\sin \theta}{\cos \theta} = 2 \][/tex]
[tex]\[ 5 \tan \theta = 2 \][/tex]
4. Solving for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{2}{5} \][/tex]
So the value of [tex]\(\tan \theta\)[/tex] is:
[tex]\[ \boxed{0.4} \][/tex]
### Part (b)
Given the equation:
[tex]\[ 5 \sin 2x = 2 \cos 2x \][/tex]
We need to solve for [tex]\(x\)[/tex] in the range [tex]\(0 \leq x < 360^\circ\)[/tex].
1. First, recall that [tex]\(\tan 2x = \frac{\sin 2x}{\cos 2x}\)[/tex]. Therefore, we can rewrite the given equation as:
[tex]\[ 5 \sin 2x = 2 \cos 2x \][/tex]
2. Dividing both sides by [tex]\(\cos 2x\)[/tex] (assuming [tex]\(\cos 2x \neq 0\)[/tex]):
[tex]\[ 5 \frac{\sin 2x}{\cos 2x} = 2 \][/tex]
[tex]\[ 5 \tan 2x = 2 \][/tex]
3. Solving for [tex]\(\tan 2x\)[/tex]:
[tex]\[ \tan 2x = \frac{2}{5} \][/tex]
4. To find the solutions for [tex]\(x\)[/tex], consider the general solution for [tex]\(\tan \phi = k\)[/tex]:
[tex]\[ \phi = \tan^{-1}(k) + n \cdot 180^\circ, \; n \in \mathbb{Z} \][/tex]
Here, [tex]\(\phi = 2x\)[/tex] and [tex]\(k = \frac{2}{5}\)[/tex].
5. Thus, the general solution for [tex]\(2x\)[/tex] is:
[tex]\[ 2x = \tan^{-1}\left(\frac{2}{5}\right) + k \cdot 180^\circ \][/tex]
6. Dividing by 2 to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{1}{2} \tan^{-1}\left(\frac{2}{5}\right) + n \cdot 90^\circ \][/tex]
7. Calculate the principal value of [tex]\(\tan^{-1}\left(\frac{2}{5}\right)\)[/tex]. It is approximately [tex]\(21.8^\circ\)[/tex]. Therefore:
[tex]\[ x = \frac{21.8^\circ}{2} + n \cdot 90^\circ \][/tex]
[tex]\[ x = 10.9^\circ + n \cdot 90^\circ \][/tex]
8. Enumerate the solutions within the given range [tex]\(0 \leq x < 360^\circ\)[/tex]:
- For [tex]\(n = 0\)[/tex], [tex]\( x = 10.9^\circ \)[/tex]
- For [tex]\(n = 1\)[/tex], [tex]\( x = 10.9^\circ + 90^\circ = 100.9^\circ \)[/tex]
- For [tex]\(n = 2\)[/tex], [tex]\( x = 10.9^\circ + 180^\circ = 190.9^\circ \)[/tex]
- For [tex]\(n = 3\)[/tex], [tex]\( x = 10.9^\circ + 270^\circ = 280.9^\circ \)[/tex]
Thus, the solutions for [tex]\(x\)[/tex] in the range [tex]\(0 \leq x < 360^\circ\)[/tex] are:
[tex]\[ \boxed{10.9^\circ, 100.9^\circ, 190.9^\circ, 280.9^\circ} \][/tex]
These are all the steps to solve the given problem correctly.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.
