# Thread: Inverse function same as original?

1. ## Inverse function same as original?

Hi guys,

Working on these last questions on my practice exam and I'm pulling my hair out...ha:

Solve the following equations in [0, 2pi]:

(a) sin^2 x + sin x = cos^2 x

I ended up with one solution, x = 3pi/2, wondering if it's ok

(b) sin 2x cos x + cos 2x sin x = 1

This one got me pretty angry but I figure there is simple solution.

I also have an aside: In my solutions manual there was a step that sort of threw me, it was about solving a trigonometric identity and they did something I hadn't seen before, they turned sin^2 x - 3 cos^2 x into sin^2 x + cos^2 x - 4 cos^2 x in order to use the pythagorean identity. So you're allowed to rewrite -3 cos^2 x as cos^2 x - 4cos^2 x? If so it's news to me...

Thanks for any help, it's much appreciated.

2. Hi

Originally Posted by DannyMath
Hi guys,

Working on these last questions on my practice exam and I'm pulling my hair out...ha:

Solve the following equations in [0, 2pi]:

(a) sin^2 x + sin x = cos^2 x

I ended up with one solution, x = 3pi/2, wondering if it's ok
Not exactly
sinē x + sin x = cosē x
sinē x + sin x = 1 - sinē x
2 sinē x + sin x - 1 = 0
which is a quadratic in sin x
sin x = -1 or sin x = 1/2
x = 3pi/2 or x = pi/6 or x = 5pi/6

Originally Posted by DannyMath
(b) sin 2x cos x + cos 2x sin x = 1

This one got me pretty angry but I figure there is simple solution.
Use sin(a+b) = sin a . cos b + sin b . cos a
with a = 2x and b = x

Originally Posted by DannyMath
I also have an aside: In my solutions manual there was a step that sort of threw me, it was about solving a trigonometric identity and they did something I hadn't seen before, they turned sin^2 x - 3 cos^2 x into sin^2 x + cos^2 x - 4 cos^2 x in order to use the pythagorean identity. So you're allowed to rewrite -3 cos^2 x as cos^2 x - 4cos^2 x? If so it's news to me...
You can factor out cosē x
cosē x - 4 cosē x = cosē x (1-4) = -3 cosē x

It is exactly the same as -3y = y - 4y with y = cosē x