sin (x+(pi/4))-sin (x-(pi/4))=1 Help to solve for solutions?

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sin (x+(pi/4))-sin (x-(pi/4))=1 Help to solve for solutions?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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one way would be to rewrite the left hand side as a single trig function it will take a raft of steps, but i think you get \[\sqrt{2}\cos(x)\]
then it should be very easy you need to use the addition angle formula, and the subtraction angle formula for sine
Sin (x+π/4)=sin x cos (π/4)+cos x sin (π/4) sin (x-π/4)=sin x cos (–π/4)-cos x sin( –π/4) Would using the sum/difference formulas help to solve? The question says they should be used but I'm not sure where to go from here.

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Other answers:

some of those are numbers right?
\[\frac{\sqrt{2}}{2}\sin(x)+\frac{\sqrt{2}}{2}\cos(x)\] is the first line
you made a mistake in the second line
sin (x-π/4)=sin x cos (–π/4)-cos x sin( –π/4) should be sin (x-π/4)=sin x cos (π/4)-cos x sin( π/4)
Yea, you're right I made a mistake with the - sign. How would I go about finding the solution?
hmm now that i look more carefully, ;perhaps you get \[\sqrt{2}\sin(x)\] when you add
no matter, now solve \[\sqrt{2}\sin(x)=1\] which is the same as \[\sin(x)=\frac{\sqrt2}{2}\]
pi/4 and 7pi/4 ?

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