How do you solve $ \sin \left( 2x \right)=-\cos \left( 2x \right) $ ?

Hint: We first simplify the equation to convert it into the form of ratio tan. Then we find the principal value of x for which $ \tan \left( 2x \right)=-1 $ . In that domain, equal value of the same ratio gives equal angles. We find the angle value for x. At the end we also find the general solution for the equation $ \tan \left( 2x \right)=-1 $ .

Complete step by step solution:
The given equation is $ \sin \left( 2x \right)=-\cos \left( 2x \right) $ . We take all the ratios on one side to get
 $ \begin{align}
  & \sin \left( 2x \right)=-\cos \left( 2x \right) \\
 & \Rightarrow \dfrac{\sin \left( 2x \right)}{\cos \left( 2x \right)}=-1 \\
 & \Rightarrow \tan \left( 2x \right)=-1 \\
\end{align} $
It’s given that $ \tan \left( 2x \right)=-1 $ . The value in is $ -1 $ . We need to find x for which $ \tan \left( 2x \right)=-1 $ .
We know that in the principal domain or the periodic value of $ -\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} $ for $ \tan x $ , if we get $ \tan a=\tan b $ where $ -\dfrac{\pi }{2}\le a,b\le \dfrac{\pi }{2} $ then $ a=b $ .
We have the value of $ \tan \left( -\dfrac{\pi }{4} \right) $ as $ -1 $ . $ -\dfrac{\pi }{2}<-\dfrac{\pi }{4}<\dfrac{\pi }{2} $ .
Therefore, $ \tan \left( 2x \right)=-1=\tan \left( -\dfrac{\pi }{4} \right) $ which gives $ 2x=-\dfrac{\pi }{4} $ .
For $ \tan \left( 2x \right)=-1 $ , the value of x is $ x=-\dfrac{\pi }{8} $ .
We also can show the solutions (primary and general) of the equation $ \tan \left( 2x \right)=-1 $ through the graph. We take $ y=\tan \left( 2x \right)=-1 $ . We got two equations $ y=\tan \left( 2x \right) $ and $ y=-1 $ . We place them on the graph and find the solutions as their intersecting points.

We can see the primary solution in the interval $ -\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2} $ is the point A as $ x=-\dfrac{\pi }{8} $ .
All the other intersecting points of the curve and the line are general solutions.


Note: Although for elementary knowledge the principal domain is enough to solve the problem. But if mentioned to find the general solution then the domain changes to $ -\infty \le x\le \infty $ . In that case we have to use the formula $ x=n\pi +a $ for $ \tan a=\tan b $ where $ -\dfrac{\pi }{2}\le a\le \dfrac{\pi }{2} $ . For our given problem $ \tan \left( 2x \right)=-1 $ , the general solution will be $ 2x=n\pi -\dfrac{\pi }{4} $ . Here $ n\in \mathbb{Z} $ . The simplified form is $ x=n\dfrac{\pi }{2}-\dfrac{\pi }{8} $ .