Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To solve the problem, we start with the given equation
[tex]\[ \sin A + \csc A = 3 \][/tex]
Recall that [tex]\(\csc A = \frac{1}{\sin A}\)[/tex]. Substituting this into the equation, we get:
[tex]\[ \sin A + \frac{1}{\sin A} = 3 \][/tex]
Let [tex]\(\sin A = x\)[/tex]. Then the equation becomes:
[tex]\[ x + \frac{1}{x} = 3 \][/tex]
Multiplying through by [tex]\(x\)[/tex], we get:
[tex]\[ x^2 + 1 = 3x \][/tex]
Rearranging the terms gives us a quadratic equation:
[tex]\[ x^2 - 3x + 1 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex]. Substituting these values in, we get:
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{5}}{2} \][/tex]
So, the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{5}}{2} \][/tex]
These solutions correspond to:
[tex]\[ \sin A = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad \sin A = \frac{3 - \sqrt{5}}{2} \][/tex]
Next, we need to evaluate [tex]\(\frac{\sin^4 A + 1}{\sin^2 A}\)[/tex] for each value of [tex]\(x\)[/tex].
Starting with the first solution, [tex]\(\sin A = \frac{3 + \sqrt{5}}{2}\)[/tex]:
[tex]\[ \frac{\left(\frac{3 + \sqrt{5}}{2}\right)^4 + 1}{\left(\frac{3 + \sqrt{5}}{2}\right)^2} \][/tex]
For the second solution, [tex]\(\sin A = \frac{3 - \sqrt{5}}{2}\)[/tex]:
[tex]\[ \frac{\left(\frac{3 - \sqrt{5}}{2}\right)^4 + 1}{\left(\frac{3 - \sqrt{5}}{2}\right)^2} \][/tex]
After solving these for the expressions, we get:
Simplifying these terms yields the results that were previously mentioned. Thus, the final solutions are derived as follows, without showing intermediate steps for the algebraic polynomials:
[tex]\[ \frac{(\sin A)^4 + 1}{(\sin A)^2} \][/tex]
For the solutions [tex]\(\sin A = \frac{3 + \sqrt{5}}{2}\)[/tex] and [tex]\(\sin A = \frac{3 - \sqrt{5}}{2}\)[/tex], the detailed stepwise solutions will lead us to the acceptable results:
[tex]\[ 1 + \left(\frac{3 + \sqrt{5}}{2}\right)^2 \quad \text{and} \quad \left(\frac{3 - \sqrt{5}}{2}^2\right) + 1 \][/tex]
[tex]\[ \sin A + \csc A = 3 \][/tex]
Recall that [tex]\(\csc A = \frac{1}{\sin A}\)[/tex]. Substituting this into the equation, we get:
[tex]\[ \sin A + \frac{1}{\sin A} = 3 \][/tex]
Let [tex]\(\sin A = x\)[/tex]. Then the equation becomes:
[tex]\[ x + \frac{1}{x} = 3 \][/tex]
Multiplying through by [tex]\(x\)[/tex], we get:
[tex]\[ x^2 + 1 = 3x \][/tex]
Rearranging the terms gives us a quadratic equation:
[tex]\[ x^2 - 3x + 1 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex]. Substituting these values in, we get:
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{5}}{2} \][/tex]
So, the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{5}}{2} \][/tex]
These solutions correspond to:
[tex]\[ \sin A = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad \sin A = \frac{3 - \sqrt{5}}{2} \][/tex]
Next, we need to evaluate [tex]\(\frac{\sin^4 A + 1}{\sin^2 A}\)[/tex] for each value of [tex]\(x\)[/tex].
Starting with the first solution, [tex]\(\sin A = \frac{3 + \sqrt{5}}{2}\)[/tex]:
[tex]\[ \frac{\left(\frac{3 + \sqrt{5}}{2}\right)^4 + 1}{\left(\frac{3 + \sqrt{5}}{2}\right)^2} \][/tex]
For the second solution, [tex]\(\sin A = \frac{3 - \sqrt{5}}{2}\)[/tex]:
[tex]\[ \frac{\left(\frac{3 - \sqrt{5}}{2}\right)^4 + 1}{\left(\frac{3 - \sqrt{5}}{2}\right)^2} \][/tex]
After solving these for the expressions, we get:
Simplifying these terms yields the results that were previously mentioned. Thus, the final solutions are derived as follows, without showing intermediate steps for the algebraic polynomials:
[tex]\[ \frac{(\sin A)^4 + 1}{(\sin A)^2} \][/tex]
For the solutions [tex]\(\sin A = \frac{3 + \sqrt{5}}{2}\)[/tex] and [tex]\(\sin A = \frac{3 - \sqrt{5}}{2}\)[/tex], the detailed stepwise solutions will lead us to the acceptable results:
[tex]\[ 1 + \left(\frac{3 + \sqrt{5}}{2}\right)^2 \quad \text{and} \quad \left(\frac{3 - \sqrt{5}}{2}^2\right) + 1 \][/tex]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.