im having trouble finding a general solution for sin2x+sinx=0
all of the examples our teacher gave in class seem much different and i'm not sure what to do. i tried to start by using the sum-to-product formula but now i'm stuck. any suggestions?
Isn't sin2x= sinx cosx?
sinx(cosx+1)=0
sin2x=2sinx cosx
sinx(2cosx+1)=0
sorry aboutthat .
To find the general solution for the equation sin(2x) + sin(x) = 0, you can follow the steps below:
Step 1: Start by using the sum-to-product formula, which states that sin(A) + sin(B) = 2*sin((A + B)/2) * cos((A - B)/2). Applying this formula to the given equation, we get:
2*sin((2x + x)/2) * cos((2x - x)/2) = 0
Simplifying this expression, we have:
2*sin(3x/2) * cos(x/2) = 0
Step 2: Now set each factor equal to zero and solve for x:
sin(3x/2) = 0 (Equation 1)
cos(x/2) = 0 (Equation 2)
Step 3: Solve Equation 1:
sin(3x/2) = 0
To find the solutions of this equation, you need to find the values of x for which sin(3x/2) = 0. Since sin(x) = 0 when x = nπ, where n is an integer, we can set 3x/2 = nπ:
3x/2 = nπ
Solving for x, we have:
x = (2nπ)/3 (Solution set for Equation 1)
Step 4: Solve Equation 2:
cos(x/2) = 0
To find the solutions of this equation, you need to find the values of x for which cos(x/2) = 0. Since cos(x) = 0 when x = (2n + 1)(π/2), where n is an integer, we can set x/2 = (2n + 1)(π/2):
x/2 = (2n + 1)(π/2)
Solving for x, we have:
x = (2n + 1)π (Solution set for Equation 2)
Step 5: Combine the solution sets from Equation 1 and Equation 2:
x = (2nπ)/3 or x = (2n + 1)π
These are the general solutions for the given equation sin(2x) + sin(x) = 0. The solutions essentially mean that x can take on any value from the solution set for Equation 1 or Equation 2. For example, if n = 0, the solutions would be x = 0 or x = π, and so on.
I hope this explanation helps you understand how to find the general solution for sin(2x) + sin(x) = 0.