Question

solve each equation on the interval

$0\le \theta \le 2\pi$

$\sin \theta -\cos \theta =-\sqrt{2}$

Short answer

$\theta =2n\pi -\frac{\pi }{4}$

Step-by-step solution

Step 1. The given information we have is

a trigonometric equation

$\sin \left(\theta \right)-\cos \left(\theta \right)=-\sqrt{2}$

Step 2. Multiplying both the side by 12

$$\begin{gathered} \frac{1}{\sqrt{2}}\sin \left(\theta \right)-\frac{1}{\sqrt{2}}\cos \left(\theta \right)=-1 \\ \frac{1}{\sqrt{2}}\cos \left(\theta \right)-\frac{1}{\sqrt{2}}\sin \left(\theta \right)=1 \\ \cos \left(\frac{\pi }{4}\right)\cos \left(\theta \right)-\sin \left(\frac{\pi }{4}\right)\sin \left(\theta \right)=\cos \left(0\right) \\ \cos \left(\theta +\frac{\pi }{4}\right)=\cos \left(0\right) \end{gathered}$$

Step 3. The general formula we get is

$\theta =2n\pi -\frac{\pi }{4}$