Question 784570
{{{2(sin(x))^2+5sin(x)=3}}}
If you make a change of variable, it gets easier to write and easier to see the solution.
Substititing {{{y=sin(x)}}}, the equation turns into
{{{2y^2+5y=3}}}
You probably know what to do with that equation.
I would do this:
{{{2y^2+5y=3}}} --> {{{2y^2+5y-3=0}}} --> {{{(2y-1)(y+3)=0}}}
From that last, factored equation, we get two solutions in terms of {{{y}}}:
{{{2y-1=0}}} --> {{{2y=1}}} --> {{{y=1/2}}} and
{{{y+3=0}}} --> {{{y=-3}}}
In terms of {{{x}}}, {{{y=sin(x)=-3}}} makes no sense.
On the other hand, from the other {{{y}}} value we get many values for {{{x}}}
{{{y=sin(x)=1/2}}}
To begin with, if we just stick to the first turn, with {{{0^o<=x<360^o}}}, we have 2 solutions: {{{highlight(30^o)}}} and {{{highlight(150^o)}}}, because
{{{sin(30^o)=1/2}}} and {{{sin(150^o)=1/2}}}
If using radians rather than degrees, for {{{0<=x<2pi}}}, the soluttions are {{{highlight(pi/6)}}} and {{{highlight(5pi/6)}}}, because
{{{sin(pi/6)=1/2}}} and {{{sin(5pi/6)=1/2}}}
If you are not restricted to the first turn, there are infinite solutions, because co-terminal angles have the same trigonometric function values, so that all the solutions could be written as
{{{(4k+1)90^o +- 60^o}}} or {{{(4k+1)pi/2 +- pi/3}}} for any integer {{{k}}}.
(Making {{{k=0}}} gives you the solutions in the first turn).