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

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

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

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marked as duplicate by DeepSea, Daniel W. Farlow, lab bhattacharjee trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 30 '17 at 3:09

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

$$\sin(x+\theta)\sin(x-\theta)=0$$

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.