Question

Double-angle Identities:
Solve this equasion for all 3 answers (leaving them in terms of pi)

2sin^2(x) - sin(x) = 1
I got it simplified this far:
-sin(x)=1-sin^2(x)
-sin(x)=cos(2x)

Where would I go from here?

Answer 1

its not right , I dont really know where u got some of the lines from

Answer 2

if anything you started with an equation with double angles, and you used identities to come up with the above

Answer 3

that first line is quadratic equation in sin(x)

Answer 4

you should solve it like a quadratic

Answer 5

(I've always been really bad at seeing the steps involved with trig, so I'm not surprised my current steps are wrong.)

Answer 6

I think I understand what you said about solving like a quadratic though, I shall try now.

Answer 7

put sin(x) = a
then you get
\[2a^2-a=1\]
\[2a^2-a-1=0\]
\[(a-1)(2a+1)=0\]
\[a=1\] or \[a=-\frac{1}{2}\]

Answer 8

now go back to trig
\[\sin(x)=1\]
\[x=\frac{\pi}{2}\]

Answer 9

\[\sin(x)=-\frac{1}{2}\]
\[x = \frac{7\pi}{6}\] or
\[x=\frac{11\pi}{6}\]