Get the best answers to your questions with the help of IDNLearn.com's experts. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.
Sagot :
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]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.
