Solve the equation sin (5/2θ) = -1/2 for 0°≤0 ≤ 180°

sin(5/2θ) = -1/2
5θ/2 = -30
θ = -12
78, 168

sin(78*2.5) = sin(195) = -.2588

So, what's wrong?

sin 210 = -1/2
5/2 θ = 210,330,...
θ = 84,132,...

when you divided by 5/2, you also need to adjust the amount you add. It's best to get a positive reference angle and then work from there.

Oh boy, we've got ourselves an equation to solve! Let's see what we can do here.

First, we have sin(5/2θ) = -1/2. To find the values of θ that satisfy this equation, we need to find where the sine of 5/2θ is equal to -1/2.

Now, we can start by finding the reference angle. The sine of a negative one-half is negative, so we know that our angle is in either the third or fourth quadrant.

To find the reference angle, we can find the inverse sine of 1/2, which is 30 degrees. Since we're dealing with a negative value, our reference angle is 180 degrees - 30 degrees = 150 degrees.

Next, let's find out what values of θ will give us a sine of -1/2. We know that in the third and fourth quadrants, the sine function is negative.

In the third quadrant, our reference angle is 150 degrees. So, we can add that to 180 degrees to get 330 degrees. This value satisfies the equation sin(5/2θ) = -1/2.

In the fourth quadrant, our reference angle is also 150 degrees. So, we can subtract that from 360 degrees to get 210 degrees. This value also satisfies the equation sin(5/2θ) = -1/2.

Therefore, we have two possible values of θ that satisfy the equation: θ = 330 degrees and θ = 210 degrees.

So, the solutions to the equation sin(5/2θ) = -1/2 for 0° ≤ θ ≤ 180° are θ = 330° and θ = 210°. Keep in mind that this equation might have other solutions outside this range, but I've only considered the values within 0° to 180°.

And there you have it, a solution with a touch of clownish charm!

To solve the equation sin(5/2θ) = -1/2 for 0°≤θ≤180°, we need to find the values of θ that satisfy this equation.

1. Start by using the inverse sine function to find θ for sin(5/2θ) = -1/2:
sin(5/2θ) = -1/2
5/2θ = -30° + k * 360° (where k is an integer)
θ = -12° + k * 72° (where k is an integer)

2. Now we need to find the values of θ that fall within the given range of 0° to 180°:
0° ≤ θ ≤ 180°

For k = 0:
θ = -12° + 0 * 72°
= -12°

For k = 1:
θ = -12° + 1 * 72°
= 60°

For k = 2:
θ = -12° + 2 * 72°
= 144°

None of these values fall within the given range of 0° to 180°.

Therefore, there are no solutions for the equation sin(5/2θ) = -1/2 in the given range.

To solve the equation sin (5/2θ) = -1/2, we need to find the values of theta (θ) that satisfy the equation in the given range of 0°≤ θ ≤ 180°.

First, we need to find the inverse sine of -1/2, which is -30°.

Since the equation contains sin(5/2θ), we can use the fact that sine has a periodicity of 360°. So we can add multiples of 360° to find all the solutions in the given range.

Let's solve the equation with the first solution:

5θ/2 = -30

Multiply both sides by 2 to isolate θ:

5θ = -60

Divide both sides by 5:

θ = -12

Therefore, one solution is θ = -12°.

Now, let's find the second solution:

Since sine has a periodicity of 360°, we can add multiples of 360° to -12° to find the next solution within the given range.

θ = -12 + 360 = 348°

So, the second solution is θ = 348°.

In the given range of 0°≤ θ ≤ 180°, the values of θ that satisfy the equation sin (5/2θ) = -1/2 are 78° and 168°.