Solve:

6sinsquared x + 5sinxcosx - 5cossquared x = 1

just cant seem to get it, any help is GREATLY APPRECIATEDD

thankuu !

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- May 29th 2009, 09:19 PMiihartheroSolve trig equation please help
Solve:

6sinsquared x + 5sinxcosx - 5cossquared x = 1

just cant seem to get it, any help is GREATLY APPRECIATEDD

thankuu ! - May 29th 2009, 09:55 PMamitface
Let $\displaystyle a=\sin x$ and $\displaystyle b=\cos x$, and move the 1 over.

Then you have $\displaystyle 6a^2+5ab-6b^2=0$. That factors nicely. See if you can work from there. - May 29th 2009, 10:14 PMiiharthero
sorry i dont no if this is a really stupid question, feeling realli thick today

but wat do u mean wen u say move the 1 over ? - May 29th 2009, 10:22 PMEnedrox
I think he actually made a mistake.

If you let $\displaystyle a=\sin x$ and $\displaystyle b=\cos x$

You get $\displaystyle 6a^2 +5ab-5b^2 = 1$ or $\displaystyle 6a^2 + 5ab - 5b^2 -1 = 0$ which do not factorise nicely. - May 29th 2009, 10:37 PMiiharthero
yea..thats wat i thought..i donno

please any one if you can help ?

thanksss - May 29th 2009, 10:44 PMEnedrox
I used a program to simplify the original equation, and it came out with:

$\displaystyle 1+11\cos 2x = 5 \sin 2x$

I tried working it out like this first:

Let's fisrt get rid of the $\displaystyle -5\cos^2 x$:

$\displaystyle 6\sin^2 x + 5\sin x\cos x = 1 + 5\cos^2 x$

$\displaystyle 6\sin^2 x + 5\sin x\cos x = 1 + 5(1-\sin^2 x)$

$\displaystyle 6\sin^2 x + 5\sin x\cos x = 1 + 5-5\sin^2 x$

$\displaystyle 11\sin^2 x + 5\sin x\cos x = 6$

$\displaystyle \sin x (11\sin x + 5\cos x) = 6$

$\displaystyle 11\sin x + 5\cos x = 6\csc x$

But it didn't give anything.

Then I found that the calculator finished up with 2x, so I gave a go to double angle formulas.

$\displaystyle 6\sin^2 + 5\sin x\cos x - 5\cos^2 x = 1$

$\displaystyle 6\sin^2 + 5\sin x\cos x - 5\cos^2 x = \sin^2 x + \cos^2 x$

$\displaystyle 5\sin^2 + 5\sin x\cos x - 5\cos^2 x = \cos^2 x$

$\displaystyle 5(\sin^2 - \cos^2 x + \sin x\cos x) = \cos^2 x$

$\displaystyle 5[-(\cos^2 - \sin^2 x) + \frac{1}{2}(2\sin x\cos x)] = \cos^2 x$

$\displaystyle 5(-\cos 2x + \frac{1}{2}\sin 2x) = \cos^2 x$

$\displaystyle -5\cos 2x + \frac{5}{2}\sin 2x = \cos^2 x$

$\displaystyle -10\cos 2x + 5\sin 2x = 2\cos^2 x$

Except I think I might have done an error (graph of step 2 and step 3 are different, dunno why) - May 29th 2009, 10:54 PMiiharthero
well thanx for taking da time to help me , wow i dont realli get wat uve done, looks extremely complicated, i dont think the question is meant to be that complicated because i can do the qs before and afta it

if it helps the solution to the question is 35 degrees 9 minutes, 120 degrees 24 minutes, 215 degrees 9 minutes and 300 degrees and 24 minutes

i jst dont no how to get there

thanx tho ! - May 29th 2009, 11:26 PMEnedrox
Well, what I've done is using the trig identities (mainly: $\displaystyle \sin^2 x + \cos^2 x = 1$)

I also used the double angle formulas:

$\displaystyle \sin(A \pm B) = \sin A\cos B \pm \sin B\cos A$

$\displaystyle \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$ - May 30th 2009, 01:48 AMs_ingram
Hi iiarthero,

the solution is attached. - May 30th 2009, 02:44 AMiiharthero
thanx for ur help !!!!!