Solve the equation sin 5/2¦È = -1/2 for 0¡ã ¡Ü 0 ¡Ü 180¡ã
I assume you mean
sin(5/2 x) = -1/2
so, 5/2 x = 210 or 330 or 570 ...
x = 84 or 132 or 228 ...
We have to stop at 180, so 84 and 132 are the solutions.
it was sin(5/2¦È) = -1/2 for 0¡ã ¡Ü 0 ¡Ü 180¡ã
it was sin(5/2¦È) = -1/2 for 0¡ã ¡Ü 0 ¡Ü 180¡ã
sin(5¦È/2)= -1/2 for 0¡ã ¡Ü 0 ¡Ü180¡ã
If positive 1/2??
To solve the equation sin (5/2π) = -1/2 for 0° ≤ θ ≤ 180°, where θ represents an angle, we need to find the value of θ that satisfies this equation.
First, let's find the reference angle, which is the smallest positive angle between the terminal side of the angle and the x-axis on the unit circle. The reference angle can be found by subtracting a multiple of 2π (360°) from the given angle, always moving in the counterclockwise direction.
The given angle is 5/2π. To find the reference angle, we subtract 2π (360°) until we obtain a value between 0 and 2π:
5/2π - 2π = 5/2π - 4π/2 = 5/2π - 2π/2 = 5/2π - 1π = (5 - 2)/2π = 3/2π
So, the reference angle is 3/2π.
Now, let's determine in which quadrants the angle is negative. In this case, sine is negative, so we look for the quadrants where sine values are negative. The sine function is negative in the third (QIII) and fourth (QIV) quadrants.
Since the reference angle is 3/2π, which lies in the second quadrant (QII), we need to find the equivalent angle in either QIII or QIV.
To find the equivalent angle in QIII, we subtract the reference angle from π (180°):
π - 3/2π = 2/2π - 3/2π = -1/2π
To find the equivalent angle in QIV, we subtract the reference angle from 2π (360°):
2π - 3/2π = 4/2π - 3/2π = 1/2π
Therefore, the two angles within the given range (0° ≤ θ ≤ 180°) that satisfy the equation sin (5/2π) = -1/2 are -1/2π and 1/2π. Note that we have expressed the angles in radians.
Please note that in the context of trigonometric equations, the solutions can also be expressed using degrees, so to convert radians to degrees, you can multiply by the conversion factor:
-1/2π * (180°/π) = -90°
1/2π * (180°/π) = 90°
Therefore, the equivalent solutions in degrees are -90° and 90°.