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Find the general solution of $\sin^2 x = \sin^2 \theta$.

My Attempt: $$\sin^2 x = \sin^2 \theta$$ $$\sin^2 x - \sin^2 \theta=0$$ $$(\sin x + \sin \theta) (\sin x - \sin \theta)=0$$ $$(2\sin \dfrac {x+\theta}{2}.\cos \dfrac {x-\theta}{2}).(2\sin \dfrac {x-\theta}{2}.\cos \dfrac {x+\theta }{2})=0$$.



marked as duplicate by DeepSea, Daniel W. Farlow, lab bhattacharjee trigonometry Jul 30 '17 at 3:09

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  • $\begingroup$ 1. What is $y$? (I think you have a typo). 2. Do you know what makes a product $0$? $\endgroup$ – Mark S. Jul 30 '17 at 1:17
  • 3
    $\begingroup$ Stop here: $(\sin x + \sin \theta) (\sin x - \sin \theta)=0$. Now, $ (\sin x + \sin \theta) = 0$ or $(\sin x - \sin \theta)=0$ $\endgroup$ – Thiago Nascimento Jul 30 '17 at 1:17

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


So either

$$ \sin(x+\theta)=0\text{ or }\sin(x-\theta)=0 $$


$$ \text{Either } x=n\pi-\theta\text{ or }x=n\pi+\theta$$

Giving the solution


As others have pointed out, you could have gotten there sooner from your third step.


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