Question 1208765
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Solve sin x + cos x = 1 for x.
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<pre>
We start from

    sin x + cos x = 1 for x.    (1)


Square both sides

    sin^2(x) + cos^2(x) + 2*sin(x)*cos(x) = 1.


Replace sin^2(x) + cos^2(x) by 1  (according the Pythagorean theorem)

    1 + 2*sin(x)*cos(x) = 1.


Cancel "1" in both sides and write in the form

    sin(2x) = 0.


Hence,  2x = {{{n*pi}}}.


It implies  x = {{{n*(pi/2)}}},   n = 0, +/-1, +/-2, . . .    (2)


We squared the original equation - so, erroneous and excessive roots could arise.
Therefore, we should check the roots (2).


Checking shows that only  x  with  n = 4k  and  n = 4k+1  satisfy the original equation.


It gives the <U>ANSWER</U> :  the solutions are  x = {{{2k*pi}}}  and  x = {{{pi/2 + 2k*pi}}},  k = 0, +/-1, +/-2, . . . 
</pre>

Solved.