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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.
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