((cosx)^2)(1+3(sinx)^2)=1

How do I find the root 54.7?

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- Jul 24th 2010, 09:44 AMStuck ManRoots
((cosx)^2)(1+3(sinx)^2)=1

How do I find the root 54.7? - Jul 24th 2010, 10:03 AMMathelogician
What do you mean by the root 54.7?

You can use the identity sinx^2+cosx^2=1 to converte sine into cosine or cosine into sine form and then solve the quadratic eqution. Just check your final solutions because of the equation's beeing in quadratic form... - Jul 24th 2010, 10:25 AMCaptainBlack
Do what Mathelogician says, giving you an equation in $\displaystyle [sin(x)]^2$, then substitute $\displaystyle u=[sin(x)]^2$ which gives you a quadratic in $\displaystyle $$ u$, solve that then ...

CB - Jul 24th 2010, 10:40 AMStuck Man
I think I have done it now. I did not get a quadratic equation. I got 2x^2-3x^4=0. I take out the factor 2x^2 and then 1-(3/2)x^2=0 gives the solutions other than 0, 180 and 360.

- Jul 24th 2010, 10:47 AMStuck Man
Is it generally preferable to let u=(sin x)^2 or let u=sin x?

- Jul 24th 2010, 10:52 AMCaptainBlack
- Jul 24th 2010, 10:53 AMMathelogician
Yes it's not quadratic and may seem a bit complicated.

Try this:

To avoid confusion let sinx = a and cosx = b and sin(2x)=c

so we have b^2+3(ab)^2=1 => 3(ab)^2=1-b^2=a^2 => (3/4)c^2 = a^2 =>

(3/4)c^2 - a^2 = 0 => (sqrt3/2 c - a)(sqrt3/2 c + a)=0

Then solve each expression; Just in solving each one, remember that sin(2x)=2sinx.cosx