(b) \( \cot x=-\frac{1}{\sqrt{15}}, \frac{3 \pi}{2}

Ask by Wilson Hills. in Malaysia

Feb 27,2025

To solve \( \cot x = -\frac{1}{\sqrt{15}} \) within the interval \( \frac{3 \pi}{2} < x < 2 \pi \), we start by recognizing that the cotangent function is negative in the fourth quadrant. Therefore, our angle \( x \) must be in this range. We know that: \[ \cot x = \frac{\cos x}{\sin x} \] Given \( \cot x = -\frac{1}{\sqrt{15}} \), we can express it in terms of sine and cosine: \[ \cos x = -\frac{1}{\sqrt{15}} \sin x \] Next, using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1\), we can substitute \( \cos x \): \[ \sin^2 x + \left(-\frac{1}{\sqrt{15}} \sin x\right)^2 = 1 \] This simplifies to: \[ \sin^2 x + \frac{1}{15} \sin^2 x = 1 \] \[ \left(1 + \frac{1}{15}\right) \sin^2 x = 1 \] \[ \frac{16}{15} \sin^2 x = 1 \] \[ \sin^2 x = \frac{15}{16} \] \[ \sin x = -\frac{\sqrt{15}}{4} \] (Because \( \sin x \) is negative in the fourth quadrant.) Now, we find \( \cos x \) using this value: \[ \cos^2 x = 1 - \sin^2 x = 1 - \frac{15}{16} = \frac{1}{16} \] \[ \cos x = \frac{1}{4} \] (We take the positive root here, since cosine is positive in the fourth quadrant.) Finally, we can find \( x \) using the reference angle \( \theta \) where \( \cot \theta = \frac{1}{\sqrt{15}} \): \[ \theta = \tan^{-1}(\sqrt{15}) \implies x = 2\pi - \theta \] This can be computed for a more exact angle if needed, but \( x \) is: \[ x = 2\pi - \tan^{-1}(\sqrt{15}) \] This gives us the desired solution for \( x \).