Hi there, I\'m working on solving the equation: 2sin^2(theta) + 3sin(theta) + 1

Question

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!

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.