Question 1196623
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Solve for all values of x in the interval [0, 2𝜋] that satisfy the equation.
4 sin(2x) = 4 cos(x)
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<pre>
4 sin(2x) = 4 cos(x)  implies, by canceling common factor "4" in both sides

    sin(2x) = cos(x),

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

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

    cos(x)*(2sin(x) - 1) = 0


So, EITHER  cos(x) = 0,  OR  sin(x) = 1/2.


If cos(x) = 0,  then  x= {{{pi/2}}}  or  x= {{{(3pi)/2}}}.


If sin(x) = 1/2,  then  x= {{{pi/6}}}  or  x= {{{pi}}} - {{{pi/6}}} = {{{(5pi)/6}}}.


So, the solution to the given equation are  {{{pi/2}}},  {{{(3pi)/2}}},  {{{pi/6}}},  {{{(5pi)/6}}}.    <U>ANSWER</U>
</pre>

Solved.