Question 1206408
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An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. 
Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
1 − 2 sin(𝜃) = cos(2𝜃)
Find the solutions in the interval [0, 2𝜋).
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        The solution in the post by @MathLover1 is  INCORRECT.

        I came to bring a correct solution.



<pre>
{{{1 -2 sin(theta) = cos(2theta)}}}

Find the solutions in the interval [{{{0}}}, {{{2pi}}}).



{{{1 -2 sin(theta) = cos(2theta)}}}....use identity {{{cos(2theta)=cos^2(theta) - sin^2(theta)}}}

{{{1 -2 sin(theta) = cos^2(theta) - sin^2(theta)}}}

{{{sin^2(theta)-cos^2(theta) -2 sin(theta) +1= 0}}}...use identity {{{cos^2(theta)=1-sin^2(theta)}}}

{{{sin^2(theta)-(1-sin^2(theta)) -2 sin(theta) +1= 0}}}

{{{sin^2(theta)-1+sin^2(theta) -2 sin(theta) +1= 0}}}

{{{2sin^2(theta) -2 sin(theta) = 0}}}...factor

{{{2sin(theta)(sin(theta) -1)  = 0}}}



If  {{{sin(theta) = 0}}}  then  {{{theta}}} = 0  or  {{{theta}}} = {{{pi}}} radians.

If  {{{sin(theta) = 1}}}  then  {{{theta}}} = {{{pi/2}}}.


<U>ANSWER</U>.  In the given interval, there are three and only three solutions, 

         x = 0,   x = {{{pi/2}}} = 1.571 radians  and   x = {{{pi}}} = 3.142 radians (rounded as requested).
</pre>

Solved (correctly)



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The post is written INCORRECTLY.


If you need the solution in the interval [0,{{{2pi}}}),
then you do not need to mention about "an arbitrary integer k".