Solve the equation sin3x+sins=0,for value of x from -180°to180°, inclusive.
sin3x+sinx = 0
since sin3x is a cubic function of sinx, that looks hard.
But your addition formula says that
sin3x + sinx = 2 sin2x cosx
So now you just have to solve
sin2x cosx = 0
2x = kπ
x = kπ + π/2
Looks like you want -180, -90, 0, 90, 180
To solve the equation sin(3x) + sin(s) = 0, for the value of x from -180° to 180°, inclusive, we can use trigonometric identities.
Step 1: Simplify the equation using the sum-to-product identity for sine function. The sum-to-product identity states that sin(A) + sin(B) = 2sin((A+B)/2)cos((A-B)/2).
sin(3x) + sin(s) = 2sin((3x+s)/2)cos((3x-s)/2) = 0
Step 2: Set each trigonometric factor equal to 0 and solve for x.
a) sin((3x + s)/2) = 0
Setting sin((3x + s)/2) = 0, we have:
(3x + s)/2 = nπ, where n is an integer
Solving for x, we get:
3x + s = 2nπ
x = (2nπ - s)/3
b) cos((3x - s)/2) = 0
Setting cos((3x - s)/2) = 0, we have:
(3x - s)/2 = (2n + 1)π/2, where n is an integer
Solving for x, we get:
3x - s = (2n + 1)π
x = (2nπ + π + s)/3
Therefore, the solutions for x are:
x = (2nπ - s)/3 and x = (2nπ + π + s)/3, where n is an integer.
Step 3: Substitute the values of x, s, and n to compute the specific solutions within the given range.
For s = -180° to 180°:
When n = -1, the solutions are:
x = (2(-1)π - s)/3 = (-2π - s)/3 = (-2π - (-180°))/3 = (-2π + 180°)/3 ≈ 60°
When n = 0, the solutions are:
x = (2(0)π - s)/3 = (-s)/3
When n = 1, the solutions are:
x = (2(1)π - s)/3 = (2π - s)/3 = (2π - (-180°))/3 = (2π + 180°)/3 ≈ 240°
Therefore, the solutions for x within the given range are approximately 60° and 240°.
To solve the equation sin(3x) + sin(x) = 0 for x from -180° to 180°, inclusive, we need to find the values of x that satisfy the equation.
First, let's combine the two terms on the left side of the equation by using the trigonometric identity:
sin(3x) + sin(x) = 2sin(x + 2x/2)cos(2x/2) (1)
Next, we simplify the equation by using the double angle identity:
2sin((3x + x)/2)cos((3x - x)/2) = 0
simplifying further, we get:
2sin(2x)cos(x) = 0 (2)
Now, we have two possibilities for the equation to be true:
1. sin(2x) = 0 and cos(x) ≠ 0
2. sin(2x) ≠ 0 and cos(x) = 0
Let's solve each possibility separately:
Possibility 1: sin(2x) = 0, cos(x) ≠ 0
To find the values of x that satisfy sin(2x) = 0, we can use the fact that sin(0) = 0.
sin(2x) = 0
2x = 0 + nπ, where n is an integer
x = 0 + (nπ/2), where n is an integer
Since we need to find the values of x from -180° to 180°, inclusive, we get the following values:
x = 0°, 90°, 180°
Possibility 2: sin(2x) ≠ 0, cos(x) = 0
To find the values of x that satisfy sin(2x) ≠ 0, we can use the fact that sin(x) = 1 or -1 when x = 90° or x = 270°, respectively.
sin(2x) ≠ 0
2x = 90° + nπ or 2x = 270° + nπ, where n is an integer
x = (90° + nπ)/2 or x = (270° + nπ)/2, where n is an integer
Since we need to find the values of x from -180° to 180°, inclusive, we get the following values:
x = -45°, 45°, 135°, 225°, 315°
Therefore, the values of x that satisfy the equation sin(3x) + sin(x) = 0 from -180° to 180°, inclusive, are:
x = -45°, 0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°.