Trig Question, Find x-intercepts

[dolpho]

Homework Statement

List all x-intercepts for y= -5sin(4x+pi/3) On the interval [-pi/6, pi/2)

Homework Equations

I know that y=sin has x-intercepts at 0, pi, and 2pi on the interval of [0, 2pi] but when I try to solve it the same way here it doesn't really come out the same.

The Attempt at a Solution

This is how I tried to solve it,

-pi/6 < 4x + pi/3 < pi/2 ----> I subtract pi/3 from both -pi/6 and pi/2

= -3pi/6 < 4x < pi/6 ----> Then I divide 4 by both

= -3pi/24 < x < pi / 24

so the x-intercepts would be -3pi/24 (-pi/8) and pi/24 and the midpoint between those but I guess its wrong :(

 

[SammyS]

Homework Statement

List all x-intercepts for y= -5sin(4x+pi/3) On the interval [-pi/6, pi/2)

Homework Equations

I know that y=sin has x-intercepts at 0, pi, and 2pi on the interval of [0, 2pi] but when I try to solve it the same way here it doesn't really come out the same.

The Attempt at a Solution

This is how I tried to solve it,

-pi/6 < 4x + pi/3 < pi/2 ----> I subtract pi/3 from both -pi/6 and pi/2

= -3pi/6 < 4x < pi/6 ----> Then I divide 4 by both

= -3pi/24 < x < pi / 24

so the x-intercepts would be -3pi/24 (-pi/8) and pi/24 and the midpoint between those but I guess its wrong :(

Hello dolpho. Welcome back to PF !

You're told that \displaystyle -\frac{\pi}{6}\le x\le\frac{\pi}{2}\ . Now find the interval on which \displaystyle 4x+\frac{\pi}{3} then occurs.

You have worked it out as if \displaystyle -\frac{\pi}{6}\le \left(4x+\frac{\pi}{3}\right)\le\frac{\pi}{2}\ .

 

[dolpho]

Hello dolpho. Welcome back to PF !

You're told that \displaystyle -\frac{\pi}{6}\le x\le\frac{\pi}{2}\ . Now find the interval on which \displaystyle 4x+\frac{\pi}{3} then occurs.

You have worked it out as if \displaystyle -\frac{\pi}{6}\le \left(4x+\frac{\pi}{3}\right)\le\frac{\pi}{2}\ .

Ahhh my mind is messed up on this problem!

I worked it another way so that

0= -5sin(4x + pi/3)

-pi/3 = 4x

-pi/12 = x

So I think that part is right but what if I wanted to find more x-intercepts along the interval of -pi / 6 and pi/2?

 

[Nessdude14]

Ahhh my mind is messed up on this problem!

I worked it another way so that

0= -5sin(4x + pi/3)

-pi/3 = 4x

-pi/12 = x

So I think that part is right but what if I wanted to find more x-intercepts along the interval of -pi / 6 and pi/2?

\Large period=\frac{2\pi}{b}
for the form:
\large asin(bx-c)+d
so your b value is 4. There will be an x intercept every half-period.

 

[SammyS]

Ahhh my mind is messed up on this problem!

I worked it another way so that

0= -5sin(4x + pi/3)

-pi/3 = 4x

-pi/12 = x

So I think that part is right but what if I wanted to find more x-intercepts along the interval of -pi / 6 and pi/2?

You could work it out:

\displaystyle -\frac{\pi}{6}\le x\le\frac{\pi}{2}\

\displaystyle -\frac{4\pi}{6}\le 4x\le\frac{4\pi}{2}\

\displaystyle -\frac{2\pi}{3}+\frac{\pi}{3}\le 4x+\frac{\pi}{3}\le2\pi+\frac{\pi}{3}\

etc.

 

[dolpho]

Ok so I think I found the answer.

0 = -5sin(4x+pi/3)

-pi/3 = 4x

-pi/12 = 1 x-int, Then I find out the period so, 2pi / 4 = pi/2

Add pi/2 to -pi/12 = 5pi/12. Then find the midpoint between those so the x-intercepts would be

-pi/12, pi/6 and 5pi/12

Anyone get the same answers?

 

[SammyS]

Ok so I think I found the answer.

0 = -5sin(4x+pi/3)

-pi/3 = 4x

-pi/12 = 1 x-int, Then I find out the period so, 2pi / 4 = pi/2

Add pi/2 to -pi/12 = 5pi/12. Then find the midpoint between those so the x-intercepts would be

-pi/12, pi/6 and 5pi/12

Anyone get the same answers?

Those are correct.