Answer :
Let's tackle the problem step-by-step.
### Part a: Solve the equation [tex]\(2 \cos^2(\theta) = 3 \sin(\theta)\)[/tex] for [tex]\(0^\circ \leq \theta \leq 360^\circ\)[/tex].
First, we'll use a trigonometric identity to simplify the equation. Recall that [tex]\(\cos^2(\theta) = 1 - \sin^2(\theta)\)[/tex]. We can rewrite the given equation using this identity:
[tex]\[ 2 \cos^2(\theta) = 2 (1 - \sin^2(\theta)) \][/tex]
So, the original equation becomes:
[tex]\[ 2 (1 - \sin^2(\theta)) = 3 \sin(\theta) \][/tex]
Expand and rearrange the equation:
[tex]\[ 2 - 2 \sin^2(\theta) = 3 \sin(\theta) \][/tex]
Move all terms to one side of the equation to set it to zero:
[tex]\[ -2 \sin^2(\theta) - 3 \sin(\theta) + 2 = 0 \][/tex]
Let's change the variable [tex]\(\sin(\theta) = x\)[/tex]:
[tex]\[ -2x^2 - 3x + 2 = 0 \][/tex]
This is a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex]. We can solve it using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = -2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 2\)[/tex].
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-2)(2)}}{2(-2)} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 16}}{-4} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{25}}{-4} \][/tex]
[tex]\[ x = \frac{3 \pm 5}{-4} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{3 + 5}{-4} = \frac{8}{-4} = -2 \][/tex]
[tex]\[ x = \frac{3 - 5}{-4} = \frac{-2}{-4} = \frac{1}{2} \][/tex]
We discard [tex]\(x = -2\)[/tex] because sine values range between -1 and 1. So, we have:
[tex]\[ \sin(\theta) = \frac{1}{2} \][/tex]
Now we find [tex]\(\theta\)[/tex] such that [tex]\(\sin(\theta) = \frac{1}{2}\)[/tex] within the given interval. The angles which satisfy this condition are:
[tex]\[ \theta = 30^\circ, 150^\circ \][/tex]
Thus, the solutions for part a) are:
[tex]\[ \theta = 30^\circ, 150^\circ \][/tex]
### Part b: Smallest positive solution of [tex]\(2 \cos^2(n\theta) = 3 \sin(n\theta)\)[/tex] given as [tex]\(10^\circ\)[/tex]. Find [tex]\(n\)[/tex] and the largest solution.
Given the smallest positive solution is [tex]\(10^\circ\)[/tex]:
[tex]\[ n\theta = 10^\circ \][/tex]
[tex]\[ \theta = \frac{10^\circ}{n} \][/tex]
To find [tex]\(n\)[/tex]:
Since the smallest positive solution is [tex]\(10^\circ\)[/tex], we have:
[tex]\[ \theta = 10^\circ \text{ implies } n \theta = n \cdot 10^\circ = 10^\circ \times 1 \][/tex]
[tex]\[ n = 1 \][/tex]
For the largest solution, since [tex]\(n = 1\)[/tex]:
The largest solution for [tex]\(n\theta\)[/tex] within the interval [tex]\(0^\circ \leq \theta \leq 360^\circ\)[/tex] will be [tex]\(360^\circ\)[/tex] when [tex]\(\theta = 360^\circ\)[/tex].
Therefore, the value of [tex]\(n\)[/tex] is:
[tex]\[ n = 1 \][/tex]
And the largest solution in the interval is:
[tex]\[ \theta = 360^\circ \][/tex]
Summarizing, for part b):
- [tex]\(n = 1\)[/tex]
- Largest solution is [tex]\(360^\circ\)[/tex]
Thus, the detailed steps yield:
1. Solutions for part a): [tex]\(\theta = 30^\circ, 150^\circ\)[/tex]
2. For part b): [tex]\( n = 1 \)[/tex] and the largest solution is [tex]\(360^\circ\)[/tex].
### Part a: Solve the equation [tex]\(2 \cos^2(\theta) = 3 \sin(\theta)\)[/tex] for [tex]\(0^\circ \leq \theta \leq 360^\circ\)[/tex].
First, we'll use a trigonometric identity to simplify the equation. Recall that [tex]\(\cos^2(\theta) = 1 - \sin^2(\theta)\)[/tex]. We can rewrite the given equation using this identity:
[tex]\[ 2 \cos^2(\theta) = 2 (1 - \sin^2(\theta)) \][/tex]
So, the original equation becomes:
[tex]\[ 2 (1 - \sin^2(\theta)) = 3 \sin(\theta) \][/tex]
Expand and rearrange the equation:
[tex]\[ 2 - 2 \sin^2(\theta) = 3 \sin(\theta) \][/tex]
Move all terms to one side of the equation to set it to zero:
[tex]\[ -2 \sin^2(\theta) - 3 \sin(\theta) + 2 = 0 \][/tex]
Let's change the variable [tex]\(\sin(\theta) = x\)[/tex]:
[tex]\[ -2x^2 - 3x + 2 = 0 \][/tex]
This is a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex]. We can solve it using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = -2\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 2\)[/tex].
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(-2)(2)}}{2(-2)} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 + 16}}{-4} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{25}}{-4} \][/tex]
[tex]\[ x = \frac{3 \pm 5}{-4} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{3 + 5}{-4} = \frac{8}{-4} = -2 \][/tex]
[tex]\[ x = \frac{3 - 5}{-4} = \frac{-2}{-4} = \frac{1}{2} \][/tex]
We discard [tex]\(x = -2\)[/tex] because sine values range between -1 and 1. So, we have:
[tex]\[ \sin(\theta) = \frac{1}{2} \][/tex]
Now we find [tex]\(\theta\)[/tex] such that [tex]\(\sin(\theta) = \frac{1}{2}\)[/tex] within the given interval. The angles which satisfy this condition are:
[tex]\[ \theta = 30^\circ, 150^\circ \][/tex]
Thus, the solutions for part a) are:
[tex]\[ \theta = 30^\circ, 150^\circ \][/tex]
### Part b: Smallest positive solution of [tex]\(2 \cos^2(n\theta) = 3 \sin(n\theta)\)[/tex] given as [tex]\(10^\circ\)[/tex]. Find [tex]\(n\)[/tex] and the largest solution.
Given the smallest positive solution is [tex]\(10^\circ\)[/tex]:
[tex]\[ n\theta = 10^\circ \][/tex]
[tex]\[ \theta = \frac{10^\circ}{n} \][/tex]
To find [tex]\(n\)[/tex]:
Since the smallest positive solution is [tex]\(10^\circ\)[/tex], we have:
[tex]\[ \theta = 10^\circ \text{ implies } n \theta = n \cdot 10^\circ = 10^\circ \times 1 \][/tex]
[tex]\[ n = 1 \][/tex]
For the largest solution, since [tex]\(n = 1\)[/tex]:
The largest solution for [tex]\(n\theta\)[/tex] within the interval [tex]\(0^\circ \leq \theta \leq 360^\circ\)[/tex] will be [tex]\(360^\circ\)[/tex] when [tex]\(\theta = 360^\circ\)[/tex].
Therefore, the value of [tex]\(n\)[/tex] is:
[tex]\[ n = 1 \][/tex]
And the largest solution in the interval is:
[tex]\[ \theta = 360^\circ \][/tex]
Summarizing, for part b):
- [tex]\(n = 1\)[/tex]
- Largest solution is [tex]\(360^\circ\)[/tex]
Thus, the detailed steps yield:
1. Solutions for part a): [tex]\(\theta = 30^\circ, 150^\circ\)[/tex]
2. For part b): [tex]\( n = 1 \)[/tex] and the largest solution is [tex]\(360^\circ\)[/tex].
