Question

Solve for x. EQUATION BELOW

Answer 1

\[\frac{ 7 }{ 6 }\sin ^2(\frac{ 7 }{ 36 }\pi x) * \cos(\frac{ 7 }{ 36 }\pi x)=0\]

Answer 2

I have it down to:

Answer 3

\[\frac{ 7 }{ 12 }\sin(\frac{ 7 }{ 18 } \pi x) * \sin(\frac{ 7 }{ 36 }\pi x) = 0\]

Answer 4

I got that by using the 1/2sin2x=sinxcosx

Answer 5

I know I can divide by the 7/12 on both sides to get rid of that. But im stuck on how to solve from there

Answer 6

\[ \sin(\frac{ 7 }{ 18 } \pi x) * \sin(\frac{ 7 }{ 36 } \pi x) = 0\]

Answer 7

you made a mistake

Answer 8

I did? where?

Answer 9

\[\frac{ 7 }{ 6 }\sin ^2(\frac{ 7 }{ 36 }\pi x) * \cos(\frac{ 7 }{ 36 }\pi x)=0\]

Answer 10

no, i'm sure i'm right to that point. I checked it with WolframAlpha.

Answer 11

1/2sin2x=sinxcosx

Answer 12

I split the sin^2 into sin*sin and then used the identity (1/2)sin2x=sinx*cosx

Answer 13

yeah i did that

Answer 14

(7/12)*(1/2)=(7/12)

Answer 15

look let me simplify this

Answer 16

and (7/36)*2=7/18

Answer 17

Ok. thanks

Answer 18

\[\sin^2(\theta)\cos(\theta)=0\]

Answer 19

it does?

Answer 20

where \[\theta = \frac{7 \pi x}{36}\]

Answer 21

I'm gonna have to show a little work. How did you get to there please?

Answer 22

That isn't the identity though.. thats for the double angle identity

Answer 23

wait

Answer 24

ok...

Answer 25

Hey phi, thanks!

Answer 26

ok the answer would be either 0, or pi/2 for the equation i put

Answer 27

i was just solving it on a piece of paper

Answer 28

now i ur case the same way

Answer 29

Oh, ok thanks mathsmind. I'm gonna see what phi says but thanks

Answer 30

sin inverse of zero is zero, and cos inverse of zero is pi/2

Answer 31

that would solve the values of x with coefficients u have

Answer 32

I would use sin^2 = 1- cos^2

\[ \frac{ 7 }{ 6 }\sin ^2(\frac{ 7 }{ 36 }\pi x) * \cos(\frac{ 7 }{ 36 }\pi x)=0 \]

we can get rid of the 7/6
\[ (1- \cos^2 x) \cos x = 0\]
or
\[ (1- \cos x) (1+ \cos x )\cos x= 0\]

now you have three equations to solve
1- cos x = 0
1+cos x =0
cos x =0

Answer 33

You lost me on that last step after you said "you can get rid of 7/6)"

Answer 34

where my x is really 7πx/36

Answer 35

multiply both sides by 6/7

Answer 36

how did it get split into three equations i mean

Answer 37

or basically how did the 1-cos^2x get split?

Answer 38

once you get to something like A * B =0 (or A*B*C=0)
you know that either A is 0 or B is 0 (or C is 0)

Answer 39

you know (a^2 - b^2) = (a-b)(a+b) right ?

Answer 40

Oh yeah and 1^2 is just 1. gotcha!

Answer 41

so I'm having an issue now solving the equations

Answer 42

work on cos ( 7πx/36)=0

that happens when the angle is pi or any multiple of pi

Answer 43

so i set (7pix/36)=pi and solve for x?

Answer 44

wait, cos is zero at pi/2 and 3pi/2
so I think you want to solve
(7pix/36)=pi/2 + n*pi

Answer 45

woah, what?

Answer 46

oh, ok i gotcha