Hi guys here is my question cos3x= sinx for the domain--- 0</ x </ Pi/2
any help would be appreciated
No it's not still easy, as you have to take into account the symmetries of the unit circle as well.
You're better off using the identity $\displaystyle \displaystyle \begin{align*} \cos{(3\theta)} = 4\cos^3{(\theta)} - 3\cos{(\theta)} \end{align*}$, then you have
$\displaystyle \displaystyle \begin{align*} \cos{(3x)} &= \sin{(x)} \\ 4\cos^3{(x)} - 3\cos{(x)} &= \sin{(x)} \\ \left[ 4\cos^3{(x)} - 3\cos{(x)} \right] ^2 &= \sin^2{(x)} \\ 16\cos^6{(x)} - 24\cos^4{(x)} + 9\cos^2{(x)} &= 1 - \cos^2{(x)} \\ 16\cos^6{(x)} - 24\cos^4{(x)} + 10\cos^2{(x)} - 1 &= 0 \\ 16X^3 - 24X^2 + 10X - 1 &= 0 \textrm{ if we let } X = \cos^2{(x)} \end{align*}$
Solve this cubic, use that to solve for x.
I don't understand why some people complicate things to such extend that their advises become useless for the students.
as I said cos3x=sin(90-3x) this is a formula of basic trigonometry no doubt about that.
then the equation is transformed to sin(90-3x) =sinx
this yields two formulas of general solutions..
1. 90-3x =(360)κ +x
2. 90-3x = (360)k+(180-x)
where k is an arbitrary integer...that is to be chosen to fit your given interval.....
no need to expand and get a qubic equation...........................