Solve the equation 3sin(2x)+3sinx=0 on the interval 0 ≤ pi <2pi. Give the sum of all the solutions.
sin 2x = 2 sin x cos c
so
6 sin x cos x + 3 sin x = 0
6 cos x + 3 = 0
cos x = - 1/2
cos is negative in quadrants 2 and 3
so pi/2 < x < 3 pi/2
180 - 60 degrees and 180 + 60 degrees
is
pi - pi/3 and pi + pi/3
2 pi/3 and 4 pi/3
don't forget the solutions at sinx = 0
To solve the equation 3sin(2x) + 3sin(x) = 0, we can start by factoring out the common term 3sin(x):
3sin(x)(sin(x) + sin(2x)) = 0
Now we need to find the values of x that make either sin(x) or (sin(x) + sin(2x)) equal to zero.
First, when sin(x) = 0:
x = 0, pi, 2pi
Next, when sin(x) + sin(2x) = 0, we can use the sum-to-product formula:
2sin(x)cos(x) + 2sin(x)cos(x) = 0
4sin(x)cos(x) = 0
This equation is true when sin(x) = 0 or cos(x) = 0.
When sin(x) = 0:
x = 0, pi, 2pi
When cos(x) = 0:
x = pi/2, 3pi/2
Therefore, the solutions on the interval 0 ≤ x < 2pi are:
x = 0, pi/2, pi, 3pi/2, 2pi
The sum of all the solutions is:
0 + pi/2 + pi + 3pi/2 + 2pi = 6pi/2 + pi/2 = 7pi/2
To solve the equation 3sin(2x) + 3sinx = 0 on the interval 0 ≤ x < 2π, we can follow the steps below:
Step 1: Simplify the equation if possible.
Using the identity sin(2x) = 2sin(x)cos(x), we can rewrite the equation as:
3(2sin(x)cos(x)) + 3sinx = 0
Simplifying further:
6sin(x)cos(x) + 3sinx = 0
Step 2: Factor out the common term.
We can factor out sin(x) from the equation:
sinx(6cos(x) + 3) = 0
Step 3: Set each factor equal to zero and solve for x.
We have two possible conditions to satisfy:
1) sinx = 0
2) 6cos(x) + 3 = 0
For the first condition, sinx = 0, the solutions on the interval 0 ≤ x < 2π are:
x = 0, π
For the second condition, 6cos(x) + 3 = 0, we need to solve for cos(x):
6cos(x) + 3 = 0
6cos(x) = -3
cos(x) = -3/6
cos(x) = -1/2
The solution for cos(x) = -1/2 on the interval 0 ≤ x < 2π is:
x = 2π/3, 4π/3
Step 4: Find the sum of all the solutions.
The solutions we obtained are: x = 0, π, 2π/3, 4π/3
To find the sum of all these solutions, simply add them:
0 + π + 2π/3 + 4π/3 = 7π/3
Therefore, the sum of all the solutions is 7π/3.