# Thread: Equations involving trig identities/formulae

1. ## Equations involving trig identities/formulae

I just need someone to guide me on how to get started on these. Generally I change the identities I spot into its equivalent, before proceeding to get lost.

Anyway, hopefully I can format these questions correctly:

a) 2sinx -cotx -cscx = 0

b) sin2x + cosx = 0

c) sinx - cosx = 0

2. Originally Posted by Cheagles
I just need someone to guide me on how to get started on these. Generally I change the identities I spot into its equivalent, before proceeding to get lost.

Anyway, hopefully I can format these questions correctly:

a) 2sinx -cotx -cscx = 0

b) sin2x + cosx = 0

c) sinx - cosx = 0
For number 1, try multiplying all by sinx, and then using $\displaystyle sin^2x+cos^2x=1$. For number two, use the double angle formula for sin2x, do you know it? Oh, and for number 3 I'd suggest multiplying by $\displaystyle (sinx+cosx)$ and see what you can do from there. Remember, you're just looking for ways to reduce to a single identity.

3. 1)

2sinx -cotx -cscx = 0

2sinx - cosx/sinx - 1/sinx =

$\displaystyle (2\sin^2 x - \cos x - 1)/sinx = 0$

$\displaystyle 1 - 2\cos^2 x - \cos x -1 = 0$

$\displaystyle 2 \cos^2x + \cos x = 0$

Now its simple

2) sin2x + cosx = 0
$\displaystyle 2\cos^2 x - 1 + \cos x = 0$
Its simple quadratic equation in cos x

3) sinx - cosx = 0

$\displaystyle \sin x = cos x$
$\displaystyle \tan x = 1$

4. Thank you, JeWiSh and arpitagarwal82. I just got done with the homework packet and ended up with only two mistakes, that I'm still trying to wrap my head around:

1) tanx + secx = 1

and

2) sin2x + cosx = 0 (I realize I already asked this, but I got it wrong. After applying the quadratic formula, I ended up with 1/2 and -1. I looked at the unit circle and got 60, 180, and 300 degrees. Where did I screw up?)

Thanks again.