Help for this equation

Question

Can anyone show me how to solve this trignometric equation :

$$\sin(2θ)-\cos(2θ)=\frac 1 2$$

Answer 1

HINT

Let use

$$\cos2\theta=\frac{1-t^2}{1+t^2}$$

$$\sin2\theta=\frac{2t}{1+t^2}$$

where

$$t=\tan \theta$$

and solve for t, that is

$$t^2+4t-3=0$$

Answer 2

It's $$\sin\left(2\theta-\frac{\pi}{4}\right)=\frac{1}{2\sqrt2}$$ and the rest is smooth: $$2\theta-\frac{\pi}{4}=(-1)^n\arcsin\left(\frac{1}{2\sqrt2}\right)+\pi n,$$ where $n$ is integer or $$\theta=\frac{\pi}{8}+ (-1)^n\frac{1}{2}\arcsin\left(\frac{1}{2\sqrt2}\right)+\frac{\pi n}{2}.$$

Answer 3

use that $$\sin(x)-\cos(y)=-2 \sin \left(-\frac{x}{2}-\frac{y}{2}+\frac{\pi }{4}\right) \sin \left(-\frac{x}{2}+\frac{y}{2}+\frac{\pi }{4}\right)$$