Find the general solution of $\sin^2 x = \sin^2 \theta$.
My Attempt: $$\sin^2 x = \sin^2 \theta$$ $$\sin^2 x - \sin^2 \theta=0$$ $$(\sin x + \sin \theta) (\sin x - \sin \theta)=0$$ $$(2\sin \dfrac {x+\theta}{2}.\cos \dfrac {x-\theta}{2}).(2\sin \dfrac {x-\theta}{2}.\cos \dfrac {x+\theta }{2})=0$$.
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In your last step you are almost there. You have
$$ (2\sin \dfrac {x+\theta}{2}.\cos \dfrac {x-\theta}{2}).(2\sin \dfrac {x-\theta}{2}.\cos \dfrac {x+\theta }{2})=0 $$
This gives
$$\sin(x+\theta)\sin(x-\theta)=0$$
So either
$$ \sin(x+\theta)=0\text{ or }\sin(x-\theta)=0 $$
Thus
$$ \text{Either } x=n\pi-\theta\text{ or }x=n\pi+\theta$$
Giving the solution
$$x=n\pi\pm\theta$$
As others have pointed out, you could have gotten there sooner from your third step.
