Please see attached image. I see what bartleby did and that makes sense but I did t a different way and got a different answer. The original question is "find all radian solutions to sinx + cosx = 0"Please show me where I went wrong ---**Problem:**\[ 43. \, ( \sin x + \cos x )^2 = (\cos^2 x - 1 - \sin^2 x) \]**Solution Steps:**1. Expand the left-hand side: \[ \sin^2 x + 2 \sin x \cos x + \cos^2 x = 0 \]2. Recognize and simplify: \[ \sin^2 x + \cos^2 x = 1 \] \[ 1 + 2 \sin x \cos x = 0 \] \[ 1 + \sin 2x = 0 \]3. Solve for \(\sin 2x\): \[ \sin 2x = -1 \]4. Solve equation: \[ 2x = \frac{3\pi}{2} + 2k\pi \] \[ x = \frac{3\pi}{4} + k\pi \]---**Explanation:**The student states:"I'm using \(\sin\), not \(\tan\), and got \(2k\pi\) instead of \(k\pi\) like Bartleby, and \(\sin\) is only \(-1\) at one angle, \(\frac{3\pi}{2}\). Please show me where I went wrong in the above calculation."This indicates the student's approach might have misunderstood the periodicity or the angles at which these functions take specific values. Verification of solutions against standard trigonometric identities or using a calculator for specific angles could clarify any discrepancies.---

Name: Please see attached image. I see what bartleby did and that makes sen
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: Please see attached image. I see what bartleby did and that makes sense but I did t a different way and got a different answer. The original question is "find all radian solutions to sinx + cosx = 0"Please show me where I went wrong ---**Problem:**