Question 937255
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Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In his post, @lwsshak3 came to conclusion "no solution: x not in given interval."


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This answer is INCORRECT. Given equation has 3 (three) solutions in the given interval.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;They are {{{pi/2}}}, {{{pi}}} and {{{3pi/2}}}. 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The reasoning in the post by @l2sshak3 is conceptually wrong, 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;it is why he missed 3 solutions.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is wrong way to teach students, so I came to bring a correct solution.



<pre>
Your starting equation is

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


Transform it step by step

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

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

    sin(2x) = 0.


Hence, 2x should be a multiple of {{{pi}}}.


Since x should be in interval ({{{0}}},{{{2pi}}}),  2x should be in interval ({{{0}}},{{{4pi}}})

    0 < 2x < {{{4pi}}}.


There are 3 possible values for 2x: {{{pi}}}, {{{2pi}}} and {{{3pi}}}.


It gives 3 possible values for x: {{{pi/2}}}, {{{pi}}} and {{{3pi/2}}}.


It is easy to check that all these 3 values satisfy given equation.


<U>ANSWER</U>.  Given equation has 3 solutions in the given interval: {{{pi/2}}}, {{{pi}}} and {{{3pi/2}}}.
</pre>

Solved correctly to teach you in a right way.