Bryant Rose
03/03/2024 · Senior High School
\( 4 \cot ^{2} x-9 \operatorname{cosec} x+6=0 \) for \( 0 \leq x \leq 2 \pi \)
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Step-by-step Solution
To solve the equation
\[
4 \cot^2 x - 9 \csc x + 6 = 0,
\]
we start by expressing \(\cot^2 x\) and \(\csc x\) in terms of \(\sin x\) and \(\cos x\):
\[
\cot^2 x = \frac{\cos^2 x}{\sin^2 x} = \frac{1 - \sin^2 x}{\sin^2 x} = \frac{1}{\sin^2 x} - 1,
\]
\[
\csc x = \frac{1}{\sin x}.
\]
Substituting these into the equation gives:
\[
4 \left( \frac{1}{\sin^2 x} - 1 \right) - 9 \frac{1}{\sin x} + 6 = 0.
\]
Multiplying through by \(\sin^2 x\) (noting that \(\sin x \neq 0\) in the interval \(0 \leq x \leq 2\pi\)) results in:
\[
4(1 - \sin^2 x) - 9 \sin x + 6 \sin^2 x = 0.
\]
This simplifies to:
\[
4 - 4 \sin^2 x - 9 \sin x + 6 \sin^2 x = 0,
\]
which can be rearranged to:
\[
(6 - 4) \sin^2 x - 9 \sin x + 4 = 0,
\]
or:
\[
2 \sin^2 x - 9 \sin x + 4 = 0.
\]
Let \(y = \sin x\). The equation becomes:
\[
2y^2 - 9y + 4 = 0.
\]
We can solve this quadratic equation using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2}.
\]
Calculating the discriminant:
\[
b^2 - 4ac = 81 - 32 = 49.
\]
Thus, we have:
\[
y = \frac{9 \pm 7}{4}.
\]
Calculating the two possible values for \(y\):
1. \(y = \frac{16}{4} = 4\) (not valid since \(\sin x\) must be in \([-1, 1]\)),
2. \(y = \frac{2}{4} = \frac{1}{2}\).
Now we have:
\[
\sin x = \frac{1}{2}.
\]
The solutions for \(\sin x = \frac{1}{2}\) in the interval \(0 \leq x \leq 2\pi\) are:
\[
x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}.
\]
Thus, the solutions to the original equation are:
\[
\boxed{\frac{\pi}{6}, \frac{5\pi}{6}}.
\]
Quick Answer
The solutions are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).
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