Question

find x where 0 ≤ x ≤ π. 2 cot²x − 5 = sec²x − tan²x ? π π , enter the smaller value first.

Explanation:

Step1: Use Trigonometric Identities

Recall the Pythagorean identities: \(\sec^{2}x - \tan^{2}x = 1\) (a fundamental trigonometric identity). So the equation \(2\cot^{2}x - 5=\sec^{2}x - \tan^{2}x\) simplifies to:
\(2\cot^{2}x - 5 = 1\)

Step2: Solve for \(\cot^{2}x\)

Add 5 to both sides of the equation:
\(2\cot^{2}x=1 + 5\)
\(2\cot^{2}x = 6\)
Divide both sides by 2:
\(\cot^{2}x=\frac{6}{2}=3\)

Step3: Solve for \(\cot x\)

Take the square root of both sides. Remember that \(\cot x=\frac{\cos x}{\sin x}\), and we are looking for solutions in the interval \(0\leq x\leq\pi\).
\(\cot x=\pm\sqrt{3}\)

Step4: Find \(x\) for \(\cot x=\sqrt{3}\)

We know that \(\cot x = \sqrt{3}\) when \(x=\frac{\pi}{6}\) (since \(\cot\frac{\pi}{6}=\sqrt{3}\))

Step5: Find \(x\) for \(\cot x = -\sqrt{3}\)

\(\cot x=-\sqrt{3}\) implies \(\tan x=-\frac{1}{\sqrt{3}}\). In the interval \(0\leq x\leq\pi\), the tangent function is negative in the second quadrant. We know that \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\), so \(x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\)

Answer:

\(\frac{1}{6}\pi\), \(\frac{5}{6}\pi\) (or \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\))