Answer :
a) To solve the equation [tex]\(2 \cos^2 \theta = 3 \sin \theta\)[/tex] for [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex], we can use trigonometric identities and algebraic techniques. Here are the steps for solving the equation:
1. Use a trigonometric identity:
Recall that [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]. Substitute this in the given equation:
[tex]\[ 2(1 - \sin^2 \theta) = 3 \sin \theta \][/tex]
2. Simplify the equation:
[tex]\[ 2 - 2 \sin^2 \theta = 3 \sin \theta \][/tex]
Move all terms to one side to form a quadratic equation in [tex]\(\sin \theta\)[/tex]:
[tex]\[ 2 \sin^2 \theta + 3 \sin \theta - 2 = 0 \][/tex]
3. Solve the quadratic equation:
Let [tex]\(x = \sin \theta\)[/tex]. The equation becomes:
[tex]\[ 2x^2 + 3x - 2 = 0 \][/tex]
Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -2\)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \][/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]
Thus, the solutions are:
[tex]\[ x = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-8}{4} = -2 \][/tex]
4. Check the validity of the solutions:
Since [tex]\(\sin \theta\)[/tex] must be in the range [tex]\([-1, 1]\)[/tex], [tex]\(x = -2\)[/tex] is not valid. So, we have:
[tex]\[ \sin \theta = \frac{1}{2} \][/tex]
5. Determine [tex]\(\theta\)[/tex] values:
For [tex]\(\sin \theta = \frac{1}{2}\)[/tex], the corresponding angles in the specified interval are:
[tex]\[ \theta = 30^{\circ}, 150^{\circ} \][/tex]
Thus, the solutions to the equation [tex]\(2 \cos^2 \theta = 3 \sin \theta\)[/tex] are:
[tex]\[ \theta = 30^{\circ}, 150^{\circ} \][/tex]
b) Given that the smallest positive solution for [tex]\(2 \cos^2 (n \theta) = 3 \sin (n \theta)\)[/tex] where [tex]\(n\)[/tex] is a positive integer is [tex]\(10^{\circ}\)[/tex], we need to find the value of [tex]\(n\)[/tex] and then determine the largest solution of the equation in the interval [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex].
1. Identify the relationship between the smallest positive solution and [tex]\(n\)[/tex]:
We previously found that without factoring [tex]\(n\)[/tex], the smallest solution to [tex]\(2 \cos^2 \theta = 3 \sin \theta\)[/tex] is [tex]\(\theta = 30^{\circ}\)[/tex]. Since the smallest positive solution of the modified equation given in the problem is [tex]\(10^{\circ}\)[/tex]:
[tex]\[ n \cdot 10^{\circ} = 30^{\circ} \][/tex]
Hence, we can solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{30^{\circ}}{10^{\circ}} = 3 \][/tex]
2. Determine the largest solution:
Knowing [tex]\(n = 3\)[/tex], we need to find [tex]\(\theta\)[/tex] such that [tex]\(0^{\circ} \leq 3 \theta \leq 360^{\circ}\)[/tex] and solve [tex]\(2 \cos^2 (3 \theta) = 3 \sin (3 \theta)\)[/tex].
As shown, the solutions for [tex]\(3 \theta\)[/tex] were:
[tex]\[ 3 \theta = 30^{\circ} \quad \text{and} \quad 150^{\circ} \][/tex]
Converting these solutions back to [tex]\(\theta\)[/tex]:
[tex]\[ \theta = 10^{\circ} \quad \text{and} \quad 50^{\circ} \][/tex]
Finally, the largest [tex]\(\theta\)[/tex] consistent with the maximum [tex]\(\theta\)[/tex] range is the largest angle in standard position within the interval:
[tex]\[ \theta = 50^{\circ} \][/tex]
Hence, the value of [tex]\(n\)[/tex] is [tex]\(3\)[/tex], and the largest solution in the interval [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex] for the equation [tex]\(2 \cos^2 (3 \theta) = 3 \sin (3 \theta)\)[/tex] is:
[tex]\[ \theta = 50^{\circ} \][/tex]
1. Use a trigonometric identity:
Recall that [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]. Substitute this in the given equation:
[tex]\[ 2(1 - \sin^2 \theta) = 3 \sin \theta \][/tex]
2. Simplify the equation:
[tex]\[ 2 - 2 \sin^2 \theta = 3 \sin \theta \][/tex]
Move all terms to one side to form a quadratic equation in [tex]\(\sin \theta\)[/tex]:
[tex]\[ 2 \sin^2 \theta + 3 \sin \theta - 2 = 0 \][/tex]
3. Solve the quadratic equation:
Let [tex]\(x = \sin \theta\)[/tex]. The equation becomes:
[tex]\[ 2x^2 + 3x - 2 = 0 \][/tex]
Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = -2\)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \][/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]
Thus, the solutions are:
[tex]\[ x = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad x = \frac{-8}{4} = -2 \][/tex]
4. Check the validity of the solutions:
Since [tex]\(\sin \theta\)[/tex] must be in the range [tex]\([-1, 1]\)[/tex], [tex]\(x = -2\)[/tex] is not valid. So, we have:
[tex]\[ \sin \theta = \frac{1}{2} \][/tex]
5. Determine [tex]\(\theta\)[/tex] values:
For [tex]\(\sin \theta = \frac{1}{2}\)[/tex], the corresponding angles in the specified interval are:
[tex]\[ \theta = 30^{\circ}, 150^{\circ} \][/tex]
Thus, the solutions to the equation [tex]\(2 \cos^2 \theta = 3 \sin \theta\)[/tex] are:
[tex]\[ \theta = 30^{\circ}, 150^{\circ} \][/tex]
b) Given that the smallest positive solution for [tex]\(2 \cos^2 (n \theta) = 3 \sin (n \theta)\)[/tex] where [tex]\(n\)[/tex] is a positive integer is [tex]\(10^{\circ}\)[/tex], we need to find the value of [tex]\(n\)[/tex] and then determine the largest solution of the equation in the interval [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex].
1. Identify the relationship between the smallest positive solution and [tex]\(n\)[/tex]:
We previously found that without factoring [tex]\(n\)[/tex], the smallest solution to [tex]\(2 \cos^2 \theta = 3 \sin \theta\)[/tex] is [tex]\(\theta = 30^{\circ}\)[/tex]. Since the smallest positive solution of the modified equation given in the problem is [tex]\(10^{\circ}\)[/tex]:
[tex]\[ n \cdot 10^{\circ} = 30^{\circ} \][/tex]
Hence, we can solve for [tex]\(n\)[/tex]:
[tex]\[ n = \frac{30^{\circ}}{10^{\circ}} = 3 \][/tex]
2. Determine the largest solution:
Knowing [tex]\(n = 3\)[/tex], we need to find [tex]\(\theta\)[/tex] such that [tex]\(0^{\circ} \leq 3 \theta \leq 360^{\circ}\)[/tex] and solve [tex]\(2 \cos^2 (3 \theta) = 3 \sin (3 \theta)\)[/tex].
As shown, the solutions for [tex]\(3 \theta\)[/tex] were:
[tex]\[ 3 \theta = 30^{\circ} \quad \text{and} \quad 150^{\circ} \][/tex]
Converting these solutions back to [tex]\(\theta\)[/tex]:
[tex]\[ \theta = 10^{\circ} \quad \text{and} \quad 50^{\circ} \][/tex]
Finally, the largest [tex]\(\theta\)[/tex] consistent with the maximum [tex]\(\theta\)[/tex] range is the largest angle in standard position within the interval:
[tex]\[ \theta = 50^{\circ} \][/tex]
Hence, the value of [tex]\(n\)[/tex] is [tex]\(3\)[/tex], and the largest solution in the interval [tex]\(0^{\circ} \leq \theta \leq 360^{\circ}\)[/tex] for the equation [tex]\(2 \cos^2 (3 \theta) = 3 \sin (3 \theta)\)[/tex] is:
[tex]\[ \theta = 50^{\circ} \][/tex]
