SOLUTION: could you PLEASE help me solve this trig identity. (6sinx)-(5cosx)-(2)=0 and the range is (0,2pi) im pretty sure i have to factor it but i cant seem to figure out how to do.

Algebra ->  Trigonometry-basics -> SOLUTION: could you PLEASE help me solve this trig identity. (6sinx)-(5cosx)-(2)=0 and the range is (0,2pi) im pretty sure i have to factor it but i cant seem to figure out how to do.      Log On


   

Question 250686: could you PLEASE help me solve this trig identity.
(6sinx)-(5cosx)-(2)=0 and the range is (0,2pi)
im pretty sure i have to factor it but i cant seem to figure out how to do.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
could you PLEASE help me solve this trig identity.
(6sinx)-(5cosx)-(2)=0
That's not an identity, it's an equation in x.
6sin - 5*sqrt(1-sin^2) - 2 = 0
6sin -2 = 5sqrt(1-sin^2)
36sin^2 - 24sin + 4 = 25 - 25sin^2
61sin^2 - 24sin - 21 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 61x%5E2%2B-24x%2B-21+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-24%29%5E2-4%2A61%2A-21=5700.

Discriminant d=5700 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--24%2B-sqrt%28+5700+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-24%29%2Bsqrt%28+5700+%29%29%2F2%5C61+=+0.815560199612356
x%5B2%5D+=+%28-%28-24%29-sqrt%28+5700+%29%29%2F2%5C61+=+-0.422117576661537

Quadratic expression 61x%5E2%2B-24x%2B-21 can be factored:
61x%5E2%2B-24x%2B-21+=+%28x-0.815560199612356%29%2A%28x--0.422117576661537%29
Again, the answer is: 0.815560199612356, -0.422117576661537. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+61%2Ax%5E2%2B-24%2Ax%2B-21+%29

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The solver uses x, it's actually sin(x)
sin(x) = 0.8155602
sin(x) = -0.4221176
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Use a calculator to find x
Email me via the thank you note to check your work, or for assistance.