How many real value(s) of x satisfy the equation
$|b| = |b \, \sin(4x)|$,
where $x \in [0, 2\pi]$ and $b$ is not zero? - HSC - SSCE Mathematics Extension 1 - Question 6 - 2024 - Paper 1
Question 6
How many real value(s) of x satisfy the equation
$|b| = |b \, \sin(4x)|$,
where $x \in [0, 2\pi]$ and $b$ is not zero?
Worked Solution & Example Answer:How many real value(s) of x satisfy the equation
$|b| = |b \, \sin(4x)|$,
where $x \in [0, 2\pi]$ and $b$ is not zero? - HSC - SSCE Mathematics Extension 1 - Question 6 - 2024 - Paper 1
Step 1
Rearrange the Equation
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Starting with the given equation, we have:
∣b∣=∣bsin(4x)∣
Since b is not zero, we can divide both sides by ∣b∣ to obtain:
1=∣sin(4x)∣
Step 2
Solve for Sin Values
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The equation ∣sin(4x)∣=1 implies:
sin(4x)=1orsin(4x)=−1
This occurs at the angles:
For sin(4x)=1: 4x=2π+2kπ, where k is an integer.
For sin(4x)=−1: 4x=23π+2kπ, where k is an integer.
Step 3
Calculate the Values of x in the Range
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
We now need to solve these for x and find the values within the interval [0,2π].
From 4x=2π+2kπ:
x=8π+2kπ
For k=0,1: x=8π,85π (within [0,2π])
From 4x=23π+2kπ:
x=83π+2kπ
For k=0,1: x=83π,87π (within [0,2π])
Combining these values gives us four distinct solutions: 8π,83π,85π,87π.
Step 4
Count the Distinct Solutions
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Thus, the total number of real values of x that satisfy the equation is 4.