Sin²(theta)=0.25
There are two possible solutions to this equation:
1) theta = 30 degrees or pi/6 radians
If we take the sine of 30 degrees, we get:
sin(30) = 0.5
Squaring this gives us:
sin²(30) = 0.25
2) theta = 150 degrees or 5pi/6 radians
If we take the sine of 150 degrees, we get:
sin(150) = -0.5
Squaring this gives us:
sin²(150) = 0.25
Therefore, the solutions are theta = 30 degrees or 150 degrees (or pi/6 radians or 5pi/6 radians).
don't forget the solutions in QIII and QIV, where sinθ = -1/2
You are correct! I apologize for my mistake.
There are actually four possible solutions to this equation:
1) theta = 30 degrees or pi/6 radians, where sin theta = 1/2
2) theta = 150 degrees or 5pi/6 radians, where sin theta = -1/2
3) theta = 210 degrees or 7pi/6 radians, where sin theta = -1/2
4) theta = 330 degrees or 11pi/6 radians, where sin theta = 1/2
Therefore, the solutions include theta = 30 degrees, 150 degrees, 210 degrees, and 330 degrees (or pi/6, 5pi/6, 7pi/6, and 11pi/6 radians).
To solve the equation sin²(theta) = 0.25, you can take the square root of both sides of the equation:
√(sin²(theta)) = ± √(0.25)
This simplifies to:
|sin(theta)| = ± 0.5
Now, we have two possibilities for the value of sin(theta):
1. sin(theta) = 0.5
2. sin(theta) = -0.5
To find the value of theta, you can use the inverse sine function (sin⁻¹) or the arcsin function. Keep in mind that there are infinite possible solutions for theta since sine is a periodic function. Let's find the solutions within the range [0, 2π] (or [0, 360°]).
For sin(theta) = 0.5:
θ1 = sin⁻¹(0.5) ≈ 30° (or π/6 radians)
θ2 = π - sin⁻¹(0.5) ≈ 150° (or 5π/6 radians)
For sin(theta) = -0.5:
θ3 = sin⁻¹(-0.5) ≈ -30° (or -π/6 radians)
θ4 = -π - sin⁻¹(-0.5) ≈ -150° (or -5π/6 radians)
Therefore, the solutions for theta in the range [0, 2π] are approximately:
θ = 30°, 150°, -30°, -150° (or π/6, 5π/6, -π/6, -5π/6 in radians).