# Trig Equation help

• Dec 8th 2008, 05:41 PM
cokeclassic
Trig Equation help
hi I have a problem,

sin^2x = 6(cosx +1)

how do i solve and get answer pi/3, /3 etc.
• Dec 8th 2008, 05:48 PM
11rdc11
Quote:

Originally Posted by cokeclassic
hi I have a problem,

sin^2x = 6(cosx +1)

how do i solve and get answer pi/3, /3 etc.

$\displaystyle 1 - \cos^2{x} = 6(\cos{x} + 1)$

$\displaystyle 1 - \cos^2{x} = 6\cos{x} + 6$

$\displaystyle 0 = \cos^2{x} + 6\cos{x} - 5$

Can you finish up from here?
• Dec 8th 2008, 05:49 PM
mr fantastic
Quote:

Originally Posted by cokeclassic
hi I have a problem,

sin^2x = 6(cosx +1)

how do i solve and get answer pi/3, /3 etc.

Start by substituting $\displaystyle \sin^2 x = 1 - \cos^2 x$.

Now re-arrange into a quadratic equation involving $\displaystyle \cos x$. Factorise and solve to get two equations for $\displaystyle \cos x$.

The only equation that leads to real solutions is $\displaystyle \cos x = 1$.
• Dec 8th 2008, 06:29 PM
cokeclassic
Ahh I had it right! I was looking at the wrong answer! (Lipssealed)
Thank you very much