Question

0=-cosx^2-sinx+sinx^2
stuck on how to solve this...

Answer 1

scratch that. lets stick witht he trig functions

Answer 2

\[0=-\cos x^2-\sin x+\sin x^2\]\[\cos x^2 = \sin x^2 - \sin x\] http://www.sosmath.com/trig/Trig5/trig5/trig5.html \[1-\sin x ^2=\sin x^2 - sin x\]

Answer 3

\[0=2\sin x^2 - \sin x-1\]

Answer 4

\[0=(2\sin x +1)(\sin x - 1)\]

Answer 5

now we have to think of the zeros.

Answer 6

are you familiar with the unit circle

Answer 7

yes! thanks so much! appreciate your help!

Answer 8

Do you still need help? At the very least I'll check your answer if you post it

Answer 9

ok. im working on it.

Answer 10

there should be 3 answers

Answer 11

ok im confused how you got from
\[2sinx ^{2}-\sin x-1\]
to 0=(2sinx+1)(sinx−1)

Answer 12

First: I used trig properties, see the website I posted, to get the equation in the same term (sin x)

Second, I set the the equation to 0 so I could factor. I could have also used the quadratic. Do you see that?

Answer 13

if x = sin x, then \[0=2x^2-x-1\]

Answer 14

solve that then you can get the zeros

Answer 15

OH. mygoodness. i didn't see that. ok. the quadratic
makes so much more sense now

Answer 16

Be careful because the quadratic just helps get it into the factored form, but because it is a trig function you need to think about the unit circle. I'll show the quadratic steps

Answer 17

yeah. with you showing the quadratic form i understand how its factored. so now i set each factor to zero and solve.
sinx=-1/2
sinx=1
so x is 7pi/6 and pi/2 right?

Answer 18

oh and 11pi/6

Answer 19

\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]\[x=\frac{1 \pm \sqrt{1-4*2*-1}}{2*2}\]\[x=\frac{1 \pm 3}{4}\]

yes good job.

Do you understand what I was saying about the quadratic just being a tool in this case and not the zeros?

Answer 20

yeah. showing how it could be factored but leaving it in the trig form. Thank you so much!

Answer 21

\[x=\frac{1 \pm 3}{4}\]\[x=\frac{-1}{2},1\]\[(\sin x + \frac{1}{2})(\sin x - 1)\]so where you would have an x you re-replace it with sinx

Answer 22

You're welcome

Answer 23

I'd print out that webpage I posted. That site has helped me so much with trig problems

Answer 24

i will! 'course, those only help if i can see the connections. thanks again!

Answer 25

It helps me to think about the problem backwards. What am i solving for? What do I need to do to get that alone? that's what I think about

Answer 26

ok. that will definitely help.