Homework SourseHi there Im working on solving the equation 2sin2theta 3sin
Hi there, I\'m working on solving the equation:
2sin^2(theta) + 3sin(theta) + 1 = 0
In the interval 0 is less than or equal to (theta) < 2pi.
I know that when factoring the above equation, we get sin(theta) = -1/2, and sin(theta) = -1.
From here, I\'m stuck:
How does sin(theta) = -1/2, and sin(theta) = -1 become (theta) = 7pi/6, 3pi/2, and 11pi/6? These are the three answers in the solution set.
Thanks in advance!
Hi there, I\'m working on solving the equation:
2sin^2(theta) + 3sin(theta) + 1 = 0
In the interval 0 is less than or equal to (theta) < 2pi.
I know that when factoring the above equation, we get sin(theta) = -1/2, and sin(theta) = -1.
From here, I\'m stuck:
How does sin(theta) = -1/2, and sin(theta) = -1 become (theta) = 7pi/6, 3pi/2, and 11pi/6? These are the three answers in the solution set.
Thanks in advance!
2sin^2(theta) + 3sin(theta) + 1 = 0
In the interval 0 is less than or equal to (theta) < 2pi.
I know that when factoring the above equation, we get sin(theta) = -1/2, and sin(theta) = -1.
From here, I\'m stuck:
How does sin(theta) = -1/2, and sin(theta) = -1 become (theta) = 7pi/6, 3pi/2, and 11pi/6? These are the three answers in the solution set.
Thanks in advance!
Solution
Hi,
As you know sin(theta) = -1/2
we need to get the theta values which will give you the values of -1/2 from 0 to 2pi (interval provided)
As we know sin(pi/6) = 1/2
but Sin is Negative in 3rd and 4th quadrants
(pi/6) + pi = 7pi/6
2pi - (pi/6) = 11pi/6
Simialrly
As you know sin(theta) = -1
we need to get the theta values which will give you the values of -1 from 0 to 2pi (interval provided)
As we know sin(pi/2) = 1
but Sin is Negative in 3rd and 4th quadrants
2pi - (pi/2) = 3pi/2
