Trig Equation help

**cokeclassic** · December 8th 2008, 05:41 PM

hi I have a problem,

sin^2x = 6(cosx +1)

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

**11rdc11** · December 8th 2008, 05:48 PM

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.

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

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

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

Can you finish up from here?

**mr fantastic** · December 8th 2008, 05:49 PM

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 $\sin^2 x = 1 - \cos^2 x$ .

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

The only equation that leads to real solutions is $\cos x = 1$ .

**cokeclassic** · December 8th 2008, 06:29 PM

Ahh I had it right! I was looking at the wrong answer!

Thank you very much

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