Question 1000865
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Solve the equation and find all real solutions:

(sinx +cosx)^2 = 1.
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Open parentheses and then simplify


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


1 + 2*sin(x)*cos(x) = 1, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<------ {{{sin^2(x) + cos^2(x)}}} = {{{1}}}


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


sin(2x) = 0, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(formula of the double argument for sine)


2x = {{{k*pi}}}, &nbsp;&nbsp;k = 0, +/- 1, +/- 2, . . . (integers)


x = {{{(k*pi)/2}}}.