Question

Trigonometry Help

@ganeshie8

Problem will be drawn out inside

Answer 1

i cant find a way to simplify the 2cosx + 1, im positive there is one but idk how

Answer 2

it is begging us to factor..

Answer 3

ik but i cant until i can find a diff way to write the 2cosx+1 right?

Answer 4

lets start and see how it goes..

Answer 5

ooooh i see, does the cosx+1 cancel out?

Answer 6

factor out further...

Answer 7

no that doesnt seem right

Answer 8

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

Answer 9

\[\large \sin x (\color{green}{2\cos x + 1}) +\color{green}{ 2\cos x + 1} = 0\]

Answer 10

\[\large (\color{green}{2\cos x + 1})( \sin x +1) = 0\]

Answer 11

see if that looks okay...

Answer 12

hm im assuming its correct just because you put it but not sure how you got it there. how does the sinx get a +1?

Answer 13

and theres 2 2cosx+1 and it becomes just 1?

Answer 14

thats a good question :) lets break it into steps

Answer 15

thanks :D

Answer 16

we have :\[\large \sin x (\color{green}{2\cos x + 1}) +\color{green}{ 2\cos x + 1} = 0\]

can we write itlike this :
\[\large \sin x (\color{green}{2\cos x + 1}) +1(\color{green}{ 2\cos x + 1}) = 0\]

?

Answer 17

yeah that works

Answer 18

oooh i see how it works now, i cant think of the word for it right now but i think i get it

Answer 19

factor by grouping is the name you are looking for?

Answer 20

yes exactly
i just wasnt seeing what it was, my confusion in my head wasnt allowing me to udnerstand

Answer 21

Good, let me put it in words :
Notice that the equation has two terms joined by a + sign. And each term has a common factor \((\color{green}{2\cos x + 1})\) which you can pull out...

Answer 22

yeah i see it perfectly now

Answer 23

\[\large \sin x (\color{green}{2\cos x + 1}) +1(\color{green}{ 2\cos x + 1}) = 0\]

pull out the GCF from both the terms, you get :

\[\large (\color{green}{2\cos x + 1}) (\sin x+1)= 0\]

Answer 24

ok just to make sure i have the correct answer, if you dont mind, id like to go through the rest of the problem with you.

and yes thank you, i see how to do that part now