Hi guys here is my question cos3x= sinx for the domain--- 0</ x </ Pi/2

any help would be appreciated

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- Oct 25th 2013, 10:43 PMGurp925Trig equation help!!
Hi guys here is my question cos3x= sinx for the domain--- 0</ x </ Pi/2

any help would be appreciated - Oct 25th 2013, 11:19 PMMINOANMANRe: Trig equation help!!
Transform the equation like this.... sin(90-3x)=sinx then the rest is easy.............

- Oct 25th 2013, 11:24 PMGurp925Re: Trig equation help!!
How did you get sin(90-3X) ?

- Oct 26th 2013, 12:23 AMProve ItRe: Trig equation help!!
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. - Oct 26th 2013, 06:08 AMMINOANMANRe: Trig equation help!!
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...........................