Question 1209453
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find all solutions:
3sin^2x+2sinx-1=0; in the interval [0,2pi)
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<pre>
Introduce new variable  u = sin(x).


Then your equation takes the form

    3u^2 + 2u - 1 = 0.


Factorize

    (3u-1)*(u+1) = 0.


Its roots are  u = 1/3  and  u= -1.


So, we consider now two cases


    (a)  sin(x) = 1/3,  which gives two solutions  x = {{{arcsin(1/3)}}}  and  x = {{{pi}}} - {{{arsin(1/3)}}}.


    (b)  sin(x) = -1,   which gives the solution   x = {{{arcsin(-1)}}}  = {{{3pi/2}}}.


At this point, the problem is solved completely.


<U>ANSWER</U>.  The solutions to the given equation in the given interval are  {{{arcsin(1/3)}}},  {{{pi}}} - {{{arsin(1/3)}}}  and  {{{3pi/2}}}.
</pre>

Solved.