Question

Sin(2x)*Sin(x)=Cos(x) ..... Solve X

Answer 1

Sin(2x)= 2Sin(x)Cos(x)

Answer 2

So I Got ..... (2Sin(x)Cos(x)Sin(x))-Cos(x)=0

Answer 3

Now What?

Answer 4

try factoring them
(2Sin(x)Cos(x)Sin(x))-Cos(x)=0

Answer 5

How?

Answer 6

@mark_o. How?

Answer 7

Cos(x) (2 Sin(x) Sin(x))-1=0

Answer 8

Cos(x) (2Sin(x)Sin(x))=1

Answer 9

hm lets try this
(2Sin(x)Cos(x)Sin(x)) = Cos(x) divide both sides by cos x
then
(2Sin(x)Sin(x)) = 1
can you continue from here?

Answer 10

@mark_o. is it possible to combine the sines?

Answer 11

yes

Answer 12

How?

Answer 13

(2Sin(x)Sin(x)) = 1 multiplying
2[Sin(x)]^2 = 1

Answer 14

oh i didn't see that 2 was not part of sine

Answer 15

ok from here
2[Sin(x)]^2 = 1
[Sin(x)]^2 = 1/2

Answer 16

ok then Sin(x)^2 = 1-Cos(2x)/2 Right?

Answer 17

nope, use power of i/2 on both sides
from
[Sin(x)]^2 = 1/2
([Sin(x)]^2)^1/2 = sqrt(1/2) correct? yes or no?

Answer 18

How would this make it solve so i get answers to the unit circle?

Answer 19

([Sin(x)]^2)^1/2 = sqrt(1/2)
\[\sin x=\sqrt{\frac{ 1 }{ 2 }}=\frac{ 1 }{ \sqrt{2} }*\frac{ \sqrt{2} }{ \sqrt{2} }=...?\]

Answer 20

X=pi/4

Answer 21

yes...you are correct
x=45deg =pi/4

Answer 22

although this seems not right there should be more to this....
because i thought....

Sin(2x)Sin(x)=Cos(x)
Sin(2x)Sin(x)-Cos(x)=0
2Sin(x)Cos(x)Sin(x)-Cos(x)=0
Cos(x)(2Sin(x)Sin(x))=1
Then Maybe.....
Cos(x)(2Sin(x)Sin(x))=1
Cos(x)=1 and 2Sin(x)=1 and Sin(x)=1

Answer 23

Attached is a example of this..

Answer 24

if it is then Cos(x)=1 has solutions 0 and 2pi
2Sin(x)=1 also as Sin(x)=1/2 solutions pi/6 and 5pi/6
Sin(x)=1 solution pi/2

Answer 25

yesssssss thats right
2Sin(x)Cos(x)Sin(x)-Cos(x)=0
could be like
Cos(x)[(2Sin(x)Sin(x))-1]=0
cos x=0 and 2Sin(x)Sin(x))-1=0
x= arc cos 0
x=90deg=pi/2
and for
2Sin(x)Sin(x))-1]=0
2Sin(x)Sin(x))=1

Answer 26

as long as restrictions are [0,2pi)

Answer 27

yes you got it.. lol..:D

Answer 28

oh that makes so much more sense after the factoring and combining of sines... lol