Question 1183233
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solve 7cos(4x)=5 for the smallest three positive solutions.
Give answer accurate to at least two decimal places as a list separated by commas.
can someone please help me better understand how to do this problem please
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<pre>
Your starting equation is 

    7cos(4x) = 5.       (1)


As you understand, it is EQUIVALENT to

    cos(4x) = {{{5/7}}}.       (2)


From (2), you have these two series of solutions

    4x = {{{arccos(5/7)}}} + {{{2pi*n}}} = 0.775193 + {{{2pi*n}}}                        (3)

and 

    4x = {{{2pi - arccos(5/7)}}} + {{{2pi*n}}} = {{{2*3.14159-0.775193}}} + {{{2pi*n}}},     (4)


where n = 0, +/-1, +/-2, +/-3, +/-4, . . . is any integer number.



From (3), dividing both its sides by 4, you have

    x = 0.193798 + {{{(pi/2)*n}}} = 0.193798 + {{{(3.14159/2)*n}}}.

    First four positive values for n= 0, 1, 2 and 3 are (rounded to 4 decimals)  x = 0.1938,  1.7646,  3.3354,  4.9062.    (5)



From (4), dividing both its sides by 4, you have

    x = 1.3770 + {{{(pi/2)*n}}} = 1.3770 + {{{(3.14159/2)*n}}}.

    First four positive values for n= 0, 1, 2 and 3 are (rounded to 4 decimals)  x = 1.3770,  2.9478,  4.5186,  6.0894.    (6)



Now your last step is to select three smallest numbers of 8 numbers of (5) and (6).


Doing this way, you obtain the  <U>ANSWER</U>  0.1938, 1.3770,  and  1.7646.
</pre>

Solved.