Why my trig equation solutions don't match the answer key

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aisha
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I have this multiplication problem it says solve 11sin theta-7=3-6sin^2 theta
algebraically for the domain 0<=theta<=2pi

I rearranged this into a quadratic formula equal to 0
6sin^2 theta+ 11 sin theta -10=0

common factored this and then got 2sin theta=0 and 3sin theta -2=0

isolating the trig function sin theta=2 this has no solution and sin theta =2/3

For this equation I got the solution to be 42 degrees and 138 degrees but this answer cannot be found in the multiple choice it says the correct answer is 222 degrees and 318 degrees. What did I do wrong??
 

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aisha said:
it says the correct answer is 222 degrees and 318 degrees.
Try plugging these angles into your original equation and see if they really do satisfy it. Then do the same with yours.
 
im not sure how to do that can someone tell me where I went wrong please
 
what do you mean? just plug in the values that you got for theta into the equation.

for example:
if your original equation is:

[tex]11sin\theta \ - \ 7 = 3 - 6sin^2 \theta[/tex]

And you want to test to see if 42 is a solution, then everytime you see a [tex]\theta[/tex] replace that with 42 and see if you get the same answer on both sides. If you do then 42 is a solution. Try doing the same for your solutions and the answers given. ( use a calculator and don't round off your answers).

I think Doc Al is trying to get you to see, that the error might not be yours :smile:
 
Last edited:
What did you say the factors of 6x^2 + 11x - 10 were?
 
my factors were

(6sin^2 theta -4 sin theta) +(15 sin theta -10) =0

2sin theta (3sin theta-2) +5 (3sin theta -2) =0

sin theta = 2 -------------and-----------sin theta=2/3
NO Solution ----------------------------reference angle =42 degrees
sin is positive in the first and second quadrants using CAST method, therefore I got the solutions to be 42 degrees, and 138 degrees.
 
Ignore the trig functions -- just work on the factoring.

You have told me that you factored

6x^2 + 11x - 10 = 0

into

2x(3x - 2) + 5(3x - 2) = 0


and then you told me that the solutions are:

x = 2 and x = 2/3.

Does that sound right to you?
 
2x(3x - 2) + 5(3x - 2) = 0

I missed a step after this

(2x+5)=0 (3x-2)=0

x=-5/2 and x= 2/3

I see one mistake I forgot the 5 before but still there is no solution for this because sin cannot=more than 1.
 
Now that we've fixed that... are you sure you've copied the problem correctly?
 
yes I just emailed my teacher she said my answer is correct :smile: thanks for all ur help