# What's the difference between these two trig identity problems?

• September 6th 2010, 04:00 AM
jayshizwiz
What's the difference between these two trig identity problems?
1) $2 \tan x=3 \sin x$
$2 \frac{\sin x}{\cos x}=3 \sin x \rightarrow$ $2 \sin x = 3 \sin x \cos x \rightarrow$ $\sin x(2 - 3 \cos x)=0$
a) $\sin x = 0 \rightarrow x =180n$
b) $\cos x = \frac{2}{3} \rightarrow x =$48.19 + 360n

2) $2 \sin 2x \cos ^2 x = \sin 2x$
$\sin 2x(2 \cos ^2 x -1) =0 \rightarrow$
a) $\sin 2x=0 \rightarrow 2x=180n \rightarrow x = 90n$
b) $\cos ^2 x = \frac{1}{2} \rightarrow \cos x = \pm \sqrt{\frac{1}{2} }$
$x = \pm$ 45 + 360n

of course the answer to number 2 is $\pm$ 45 + 180n
but the answer to number 1 is 48.19 + 360n! So what's the difference between the two? They are both cosine functions. I just thought always add 360 degrees because that's when the angle repeats itself.

(*fixed the bold...)
• September 6th 2010, 08:04 AM
Pim
$\cos x = \pm \frac{1}{2}\sqrt{2}$ has four solutions on $[0,360]$
These are 45, 135, 225 and 315
45 + 360n and 135 + 360n can be rewritten as 45+180n, because they are half a period apart.
Same goes for 135 and 315 (where -45 is 315-360)
You could sum up all solutions as 45+90n

Do you understand now?
• September 6th 2010, 09:03 AM
jayshizwiz
touche