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).
Sin²(theta)=0.25
3 answers
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).
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).