# I have a problem to calculate this kind of quadratic equation 5sinsquarex+sin2x-3=0. I know that if we have only sinx as the second part of this equation, it'll be easy to find roots: it would be...

I have a problem to calculate this kind of quadratic equation 5sinsquarex+sin2x-3=0. I know that if we have only sinx as the second part of this equation, it'll be easy to find roots: it would be sinx= (-b+squarerootof bsquare-4ac)all over 2a. but How can I find sin2x? thank you very much

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

`5sin^2(x)+sin(2x)-3=0`

We can write this equation as

`5sin^2(x)+2sin(x)cos(x)-3=0`

Let sin(x)=t

so   `cos(x)=sqrt(1-sin^2(x))=sqrt(1-t^2)`

So equation can be written as

`5t^2+2tsqrt(1-t^2)-3=0`

`2tsqrt(1-t^2)=3-5t^2`

squaring both side

`4t^2(1-t^2)=9+25t^4-30t^2`

`4t^2-4t^4-25t^4-9+30t^2=0`

`-29t^4+34t^2-9=0`

`29t^4-34t^2+9=0`

`t^2=(34+-sqrt(34^2-4xx9xx29))/(2xx29)`

`t^2=(34+-sqrt(112))/58`

`t^2=(34+-10.583)/58`

`t^2=.76867`

`t=+-.8767`

`sin(x)=+-.8767`

`x=+-61.25^0`

`or`

`t^2=.40374`

`t=+-.6354`

`sin(x)=+-.6354`

`x=+-39.45^0`

Substitute the values of x in original equation to find the value of x.It may possible all value of  above x not satisfy the give equation.

Once you have x the you can calculate sin(2x).