Question 1188985
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3sin(2x−6)−1=0
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<pre>
We transform given equation this way

    sin(2x-6) = 1/3


After that, we get two possible forms

         (a)  2x-6 = arcsin(1/3) + 2k*pi

    and

         (b)  2x-6 = pi - arcsin(1/3) + 2k*pi


It gives two expressions for x

         (a)  x = (arcsin(1/3) + 6)/2 + k*pi
        
    and

         (b)  x = (pi - arcsin(1/3) +6)/2 + k*pi


Giving values  k= 0 and -1 for(a)  and  k=0 and -1 for (b), we get 4 values for x in the interval [0,2pi), in radians


    1)  x = {{{(6+arcsin(1/3))/2}}} = {{{(6+0.3398)/2}}} = 3.169918455,


    2)  x = {{{(6+arcsin(1/3))/2}}} - {{{pi}}} = {{{(6+0.3398)/2}}} - {{{3.14159}}} = 0.028328455,


    3)  x = {{{(6 + pi-arcsin(1/3))/2}}} = {{{(6+3.14159-0.3398)/2}}} = 4.400876545,


    4)  {{{(6 + pi-arcsin(1/3))/2}}} - {{{pi}}} = {{{(6+3.14159-0.3398)/2}}} -  {{{3.14159}}} = 1.259286545.


The minimum of these 4 values is  0.028328455.     <U>ANSWER</U>
</pre>

Solved.