Question 1182204
.


This problem was posted to the forum couple of days ago,  and I gave its full solution under this link


<A HREF=https://www.algebra.com/algebra/homework/Functions/Functions.faq.question.1182143.html>https://www.algebra.com/algebra/homework/Functions/Functions.faq.question.1182143.html</A>


https://www.algebra.com/algebra/homework/Functions/Functions.faq.question.1182143.html



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;For your convenience, &nbsp;I &nbsp;copy-paste that solution here again



<pre>
Starting from

    sin(x) + cos(x) = 1,      (1)


square both sides

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



Take into account that  {{{sin^2(x)}}} + {{{cos^2(x)}}} == 1.


You will get then from  (2)


    2*sin(x)*cos(x) = 0,

or

      sin(x(*cos(x) = 0.


So, equation (1) implies

    sin(x) = 0  or  cos(x) = 0.


It gives  x = {{{k*pi}}}  or  x = {{{pi/2 + k*pi}}},   where k is any integer number.


Checking these potential solutions,  we get the final answers  x = {{{2pi*n}}},  or  x = {{{pi/2 + 2pi*n}}},  where n is any integer number.


Of the given four optional choices,  only  x= 90°  is the solutions to the given equation.
</pre>

Solved.