Question 1001690
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{{{2*sin^2(x)}}} = {{{3*sin(x)}}} + {{{5}}}.


Let us introduce a new variable,  y = sin(x).

Then the equation takes the form


{{{2y^2}}} = {{{3y + 5}}},     or


{{{2y^2}}} - {{{3y}}} - {{{5}}} = {{{0}}}.


Solve it by using the quadratic formula:


{{{y[1,2]}}} = {{{(3 +- sqrt(3^2 + 4*2*5))/4}}} = {{{(3 +- sqrt(49))/4}}} = {{{(3 +- 7)/4}}}.


{{{y[1]}}} = -1   ----->   sin(x) = -1   ----->   x = {{{3*pi/2}}}.


{{{y[2]}}} = {{{10/4}}}  doesn't fit,  because  sin(x)  can not be more than  1.


<U>Answer</U>. &nbsp;x = {{{3*pi/2}}}.