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To determine the solutions of the equation [tex]\(\cot^2 x + 2 = 2 \csc x\)[/tex] within the interval [tex]\([0, 2\pi)\)[/tex], follow these steps:
1. Rewrite the Equation:
[tex]\[ \cot^2 x + 2 = 2 \csc x \][/tex]
Recall that [tex]\(\cot x = \frac{\cos x}{\sin x}\)[/tex] and [tex]\(\csc x = \frac{1}{\sin x}\)[/tex]. Rewriting in terms of sine and cosine:
[tex]\[ \left(\frac{\cos x}{\sin x}\right)^2 + 2 = 2 \frac{1}{\sin x} \][/tex]
2. Simplify the Trigonometric Expressions:
[tex]\[ \frac{\cos^2 x}{\sin^2 x} + 2 = \frac{2}{\sin x} \][/tex]
Multiply through by [tex]\(\sin^2 x\)[/tex] to clear the denominators:
[tex]\[ \cos^2 x + 2 \sin^2 x = 2 \sin x \][/tex]
Recall that [tex]\(\cos^2 x = 1 - \sin^2 x\)[/tex], substitute this into the equation:
[tex]\[ 1 - \sin^2 x + 2 \sin^2 x = 2 \sin x \][/tex]
Simplify:
[tex]\[ 1 + \sin^2 x = 2 \sin x \][/tex]
3. Form a Quadratic Equation:
Let [tex]\(u = \sin x\)[/tex]. Then:
[tex]\[ 1 + u^2 = 2u \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ u^2 - 2u + 1 = 0 \][/tex]
Solve the quadratic equation using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ u = \frac{2 \pm \sqrt{4 - 4}}{2} = \frac{2 \pm 0}{2} = 1 \][/tex]
4. Solve for [tex]\(x\)[/tex]:
Since [tex]\(\sin x = 1\)[/tex], we need to find the values of [tex]\(x\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex]:
[tex]\[ \sin x = 1 \implies x = \frac{\pi}{2} \][/tex]
5. Check the Provided Choices:
The possible solutions to the equation are where [tex]\(\sin x = 1\)[/tex]:
- [tex]\(\frac{3\pi}{2}\)[/tex] is not valid because [tex]\(\sin (\frac{3\pi}{2}) = -1\)[/tex]
- [tex]\(0, \pi\)[/tex] are not valid because [tex]\(\sin 0 = 0\)[/tex] and [tex]\(\sin \pi = 0\)[/tex]
- [tex]\(\frac{\pi}{2}, \frac{3\pi}{2}\)[/tex] results in [tex]\(\frac{\pi}{2}\)[/tex] being valid and [tex]\(\frac{3\pi}{2}\)[/tex] invalid
- [tex]\(\frac{\pi}{2}\)[/tex] is valid
Therefore, the correct solutions from the given choices are:
[tex]\[ \frac{\pi}{2}, \frac{\pi}{2} \][/tex]
This corresponds to [tex]\(\boxed{\frac{\pi}{2}}\)[/tex].
1. Rewrite the Equation:
[tex]\[ \cot^2 x + 2 = 2 \csc x \][/tex]
Recall that [tex]\(\cot x = \frac{\cos x}{\sin x}\)[/tex] and [tex]\(\csc x = \frac{1}{\sin x}\)[/tex]. Rewriting in terms of sine and cosine:
[tex]\[ \left(\frac{\cos x}{\sin x}\right)^2 + 2 = 2 \frac{1}{\sin x} \][/tex]
2. Simplify the Trigonometric Expressions:
[tex]\[ \frac{\cos^2 x}{\sin^2 x} + 2 = \frac{2}{\sin x} \][/tex]
Multiply through by [tex]\(\sin^2 x\)[/tex] to clear the denominators:
[tex]\[ \cos^2 x + 2 \sin^2 x = 2 \sin x \][/tex]
Recall that [tex]\(\cos^2 x = 1 - \sin^2 x\)[/tex], substitute this into the equation:
[tex]\[ 1 - \sin^2 x + 2 \sin^2 x = 2 \sin x \][/tex]
Simplify:
[tex]\[ 1 + \sin^2 x = 2 \sin x \][/tex]
3. Form a Quadratic Equation:
Let [tex]\(u = \sin x\)[/tex]. Then:
[tex]\[ 1 + u^2 = 2u \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ u^2 - 2u + 1 = 0 \][/tex]
Solve the quadratic equation using the quadratic formula [tex]\(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ u = \frac{2 \pm \sqrt{4 - 4}}{2} = \frac{2 \pm 0}{2} = 1 \][/tex]
4. Solve for [tex]\(x\)[/tex]:
Since [tex]\(\sin x = 1\)[/tex], we need to find the values of [tex]\(x\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex]:
[tex]\[ \sin x = 1 \implies x = \frac{\pi}{2} \][/tex]
5. Check the Provided Choices:
The possible solutions to the equation are where [tex]\(\sin x = 1\)[/tex]:
- [tex]\(\frac{3\pi}{2}\)[/tex] is not valid because [tex]\(\sin (\frac{3\pi}{2}) = -1\)[/tex]
- [tex]\(0, \pi\)[/tex] are not valid because [tex]\(\sin 0 = 0\)[/tex] and [tex]\(\sin \pi = 0\)[/tex]
- [tex]\(\frac{\pi}{2}, \frac{3\pi}{2}\)[/tex] results in [tex]\(\frac{\pi}{2}\)[/tex] being valid and [tex]\(\frac{3\pi}{2}\)[/tex] invalid
- [tex]\(\frac{\pi}{2}\)[/tex] is valid
Therefore, the correct solutions from the given choices are:
[tex]\[ \frac{\pi}{2}, \frac{\pi}{2} \][/tex]
This corresponds to [tex]\(\boxed{\frac{\pi}{2}}\)[/tex].
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