How Do You Isolate sin a in a Quadratic Trigonometric Equation?

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To isolate sin a in the quadratic trigonometric equation 6sin^2 a - 11sin a + 4 = 0, it is suggested to substitute sin a with a variable, such as y, transforming the equation into 6y^2 - 11y + 4 = 0. This allows for easier manipulation and solving of the quadratic equation. Once the values for y are found, they can be substituted back to find the corresponding angles for sin a. The domain for the solutions is specified as -360 to 360 degrees. This method simplifies the process of isolating sin a and finding the necessary angles for the exam.
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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!
 
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ditto_299 said:
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.
 
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