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

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

thankuu !

2. 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.

3. 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 ?

4. 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.

5. yea..thats wat i thought..i donno
please any one if you can help ?
thanksss

6. 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)

7. 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 !

8. 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:
$\displaystyle \sin(A \pm B) = \sin A\cos B \pm \sin B\cos A$
$\displaystyle \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$

9. Hi iiarthero,

the solution is attached.

10. thanx for ur help !!!!!