Question 1155029
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<pre>

The general statement is 


    If cos(a) + cos(b) = 0, then


        EITHER  a + b = {{{pi}}} + {{{2k*pi}}}

        OR      a - b = {{{pi}}} + {{{2k*pi}}}.


If apply it to  x  and  3x,  then


    EITHER  x + 3x = {{{pi}}} + {{{2k*pi}}}

    OR      |x - 3x| = {{{pi}}} + {{{2k*pi}}}.


First case gives

    4x = {{{(2k+1)*pi}}};  hence,  x = {{{pi/4}}},  or  x = {{{3pi/4}}},  or  x = {{{5pi/4}}},  or  x = {{{7pi/4}}}.


Second case gives

   2x = {{{(2k+1)*pi}}};  hence,  x = {{{pi/2}}},  or  x = {{{3pi/2}}}.


These 6 listed values are the full set of solutions to the given equation in given interval.
</pre>

Solved.



Another approach is possible.


<pre>
Use the general formula


    cos(a) + cos(b) = {{{2*cos((a+b)/2)*cos((a-b)/2)}}}.


When you apply it with  x  and  3x, you get


    cos(x) + cos(3x) = 2*cos(2x)*cos(x) = 0,


which leads you to the SAME answer.
</pre>

Solved.