# Trig Question, Find x-intercepts

## 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 :(

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SammyS
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## 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 $\displaystyle -\frac{\pi}{6}\le \left(4x+\frac{\pi}{3}\right)\le\frac{\pi}{2}\ .$

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?

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
Staff Emeritus
Homework Helper
Gold Member
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.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
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