Question

HELP ME SOLVE THIS.
sin(2 theta)-cos(theta)-2 sin(theta)+1=0

Answer 1

I know you can use the double angle formula and whatnot, but im still baffled. Also, it asks to find the values only in 0 to 2pi.

Answer 2

please don't give the answer. Just show me clues to solving this.

Answer 3

What would you recommend?

Answer 4

This is precalculus.

Answer 5

use the trig identity sin(2 theta)=2sin(theta)cos(theta)

Answer 6

i did

Answer 7

im still stumped. mothing can be factored after

Answer 8

\[\large 2\sin \theta \cos \theta -\cos \theta-2 \sin \theta +1=0\]
from wolfram looks like you have to use the half angle formulas to solve this... see "alternate forms" http://www.wolframalpha.com/input/?i=sin(2+theta)-cos(theta)-2+sin(theta)%2B1%3D0

Answer 9

You used wolfram to solve it?

Answer 10

How may I use wolfram to study better?

Answer 11

I tried to use it but there were so many different calcluators.

Answer 12

No i'm saying look at the alternate forms, as a way to help solve it. It is not at all obvious, though.

Answer 13

I also want to see the steps wolfram used to solve this. All I get is the final outcome. I want to learn how to derive it.

Answer 14

Hey look...\[\large 2\sin \theta \cos \theta -\cos \theta-2 \sin \theta +1=0\] \[\large \cos \theta(2\sin \theta -1)-1(2 \sin \theta -1)=0\]

Answer 15

\[\large (\cos \theta-1)(2\sin \theta -1)=0 \]

Answer 16

My goodness. You are brilliant.

Answer 17

I got to your first step, but I didn't picture the negative 1 as being a factor..

Answer 18

Haha thank you

Answer 19

How did you think of that?

Answer 20

Tried factoring cosine out of the first two terms, noticed the similarity with the remaining terms

Answer 21

Sometimes trial and error pays off.

Answer 22

2sin(theta)cos(theta)-cos(theta)-2sin(theta)+1

Then,
Cos(theta)*(2sin(theta)-1)-2sin(theta)+1

Then,
(cos(theta) +1) (2sin(theta)-1) = 0

For my solution, I got pi, pi/6/ and 5pi/6. However my friend said it was 0 instead of pi. Can someone show my error?

Answer 23

Ahh, i distributed it backwards, and found my error. Thank you.