Solving 6sin^2 a - 11sin a + 4 = 0: Help Needed!

[ditto_299]

So, I know how to solve them in general, but this one in particular, what do you do when the sin or cos etc is squared?

6sin^2 a - 11sin a + 4 = 0 , domain is from -360 until 360 degrees

we learned that, you need to isolate sin a, and then find the possible angles. I just need to know how to isolate sin a.

so i thought maybe this...

6 (sin a + sin a) - 11sin a = -4

but i don't know from here on...
please help! exam is coming up!

 

[HallsofIvy]

So, I know how to solve them in general, but this one in particular, what do you do when the sin or cos etc is squared?

6sin^2 a - 11sin a + 4 = 0 , domain is from -360 until 360 degrees

we learned that, you need to isolate sin a, and then find the possible angles. I just need to know how to isolate sin a.

so i thought maybe this...

6 (sin a + sin a) - 11sin a = -4

What? sin^2 a is (sin a)(sin a) NOT "sin a+ sin a"?

but i don't know from here on...
please help! exam is coming up!

Perhaps it would be simpler to replace sin a by y:
6y^2- 11y+ 4= 0. Can you solve that quadratic equation? After finding solutions, that is, specific values for y, you can solve sin a= y.

 

[Architeuthis Dux]

I would suggest approaching this problem by using the quadratic formula. Since the equation contains a squared term, it is a quadratic equation in terms of sin a. Therefore, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, our "x" will represent sin a. So, we can plug in our values for a, b, and c from the given equation:

a = 6
b = -11
c = 4

Plugging these values into the formula, we get:

sin a = (-(-11) ± √((-11)^2 - 4(6)(4))) / 2(6)

simplify:

sin a = (11 ± √(121 - 96)) / 12

sin a = (11 ± √25) / 12

sin a = (11 ± 5) / 12

sin a = 4/12 or 16/12

sin a = 1/3 or 4/3

Now, we need to consider the domain given in the problem (-360 to 360 degrees). We know that the sine function repeats itself every 360 degrees, so we can add or subtract 360 from our solutions to get all possible angles within the given domain.

Therefore, our solutions for sin a are:

sin a = 1/3, 4/3, 361/3, 364/3, 721/3, 724/3

Since we are only interested in angles within the given domain, we can disregard the last two solutions (361/3 and 364/3). So, our final solutions for sin a are:

sin a = 1/3, 4/3, 721/3, 724/3

I hope this helps you understand how to isolate sin a in a quadratic equation and find the possible angles within a given domain. Good luck on your exam!