Solve for x: (sin x)^2 - (cos x)^2 + sin x = 0

Question

zepdrix · Answer

$$\huge \sin^2 x - \cos^2 x+\sin x = 0$$Hmm, we'll use a familiar identity to change the cosine to a SINE term.$$\large \sin^2 x + \cos^2 x = 1 \qquad \rightarrow \qquad \cos^2x=(1-\sin^2x)$$
Using this identity gives us,$$\huge \sin^2x-(1-\sin^2x)+\sin x=0$$Which simplifies to $$\huge 2\sin^2 x+\sin x-1=0$$

zepdrix · Answer

From here we can use the quadratic formula to solve for sine x! :D
Understand the steps we did so far?

anonymous · Answer

Thanks! Yep you used an identity right?
So do I factor it to (2sinx+1)(sinx -1)? Not sure

anonymous · Answer

can't solve for x, I need to find my point of inflection..

zepdrix · Answer

(2sinx-1)(sinx+1)=0
Hmm I think it factors like this actually :) I think your signs are backwards.
Hah thats funny, i was gonna jump right to the quadratic formula... I wasn't able to see the factors lol.

zepdrix · Answer

Points of inflection..? :o whu?

anonymous · Answer

For calculus.. I got to this step from taking the 2nd derivative but couldn't solve for x to get my point of inflection.

zepdrix · Answer

Oh i see :)

anonymous · Answer

Hehe :D and thank you!

zepdrix · Answer

Oh you got it from here? :) yay team \:D/

anonymous · Answer

(2sinx-1)(sinx+1)=0 
I multiplied this out to check but it doesn't equal to 2(sinx)^2 + sinx−1=0.
Instead it equals to 2(sinx)^2 - sinx−1=0

So gotta do the quadratic formula? Ahh.

zepdrix · Answer

2sin^2 +2sinx -sinx - 1
that's what it comes out to i think.

zepdrix · Answer

Make sure you're using the factors that I posted, not the one you wrote out earlier :) you had your signs backwards.

anonymous · Answer

Okay I got x= pi/6 and 3pi/2

zepdrix · Answer

Hmm yah those looks good.
Is 5pi/6 also a possibility? Hopefully I'm not thinking about that incorrectly :3

anonymous · Answer

Oh yeah it is. Thanks for reminding. Oh my