Question

find x where 0 ≤ x ≤ 2π. sin x − cot²x = 1 \frac{π}{?}

Explanation:

Step1: Rewrite using trigonometric identity

Recall that \(\cot^{2}x=\csc^{2}x - 1\) (from the Pythagorean identity \(1+\cot^{2}x=\csc^{2}x\)). Substitute this into the equation:

$$ \sin x-(\csc^{2}x - 1)=1 $$

Simplify the equation:

$$ \sin x-\csc^{2}x + 1=1 $$

Subtract 1 from both sides:

$$ \sin x-\csc^{2}x=0 $$

Since \(\csc x=\frac{1}{\sin x}\), we can rewrite \(\csc^{2}x\) as \(\frac{1}{\sin^{2}x}\):

$$ \sin x-\frac{1}{\sin^{2}x}=0 $$

Multiply through by \(\sin^{2}x\) (note that \(\sin x
eq0\) because if \(\sin x = 0\), \(\csc x\) is undefined) to get:

$$ \sin^{3}x-1 = 0 $$

Step2: Solve the cubic equation

We have the equation \(\sin^{3}x-1 = 0\), which can be factored as a difference of cubes: \((\sin x - 1)(\sin^{2}x+\sin x + 1)=0\).

First, consider \(\sin x-1 = 0\), so \(\sin x=1\). The solution to \(\sin x = 1\) in the interval \(0\leq x\leq2\pi\) is \(x=\frac{\pi}{2}\).

Now, consider the quadratic factor \(\sin^{2}x+\sin x + 1 = 0\). The discriminant of this quadratic in terms of \(\sin x\) is \(\Delta=1^{2}-4\times1\times1=1 - 4=- 3<0\). Since the discriminant is negative, there are no real solutions for \(\sin x\) from this quadratic factor.

We also need to check the domain restriction we had earlier ( \(\sin x
eq0\) ). The solution \(x = \frac{\pi}{2}\) has \(\sin x=1
eq0\), so it is valid.

Answer:

The value of \(x\) in the interval \(0\leq x\leq2\pi\) that satisfies the equation is \(\frac{\pi}{2}\), so the denominator in \(\frac{\pi}{?}\) is \(2\).