Question 1208069
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what {{{highlight(cross(is))}}} <U>are</U> the possible values of x in 2sin^2(x-30)=cos 60
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<pre>
They want you find all real solutions to this equation

    2sin^2(x-30) = cos(60).    (1)


We have 

    cos(60) = 1/2;  so, this equation takes the form

    2sin^2(x-30) = 1/2.



Simplify it step by step

    sin^2(x-30) = 1/4,

    sin(x-30) = +/-{{{sqrt(1/4)}}}},

    sin(x-30) = +/- 1/2.    (2)



From this equation, there are two infinite sets of roots for  x-30  to equation  (2)

    x-30 = {{{30 + k*pi}}},  k = 0, +/-1, +/-2, . . . 

    x-30 = {{{150 + k*pi}}},  k = 0, +/-1, +/-2, . . . 



It gives two infinite sets of roots for  x  to equation  (1)

    x = {{{60 + k*pi}}},  k = 0, +/-1, +/-2, . . . 

    x = {{{k*pi}}},  k = 0, +/-1, +/-2, . . . 



<U>ANSWER</U>.  Equation (1) has two infinite sets of solutions

             x = {{{60 + k*pi}}},  k = 0, +/-1, +/-2, . . . 

             x = {{{k*pi}}},  k = 0, +/-1, +/-2, . . . 
</pre>

Solved.



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To tutor @greenestamps and to all other readers of this post.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Thank you for your attention, &nbsp;but my solution, &nbsp;as it is/was written by me 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;from the very beginning, &nbsp;is correct and includes all the cases &nbsp;&nbsp;sin(x-30) = +/- 1/2.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is because the general formula  uses not  full period  &nbsp;{{{2pi}}},  &nbsp;but  half period &nbsp;{{{pi}}},  &nbsp;which absorbs/covers all these cases.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;So, everything is and was correct in my post. &nbsp;&nbsp;There is no reason to worry.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;This alarm from @greenestamps is false.