SOLUTION: Please help me to solve : 2 tan y + 5 cosy y = 0 for 0 < y < 6 Thank You.

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Question 252522: Please help me to solve :
2 tan y + 5 cosy y = 0 for 0 < y < 6
Thank You.
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
2 tan y + 5 cosy y = 0 for 0 < y < 6
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Do you mean 2 tan y + 5 cosy = 0 ??
2sin/cos + 5cos = 0
2sin + 5cos^2 = 0
2sin +5(1-sin^2) = 0
-5sin^2 + 2sin + 5 = 0
Now it's a quadratic in sin(y)
Sub x for sin(y):
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -5x%5E2%2B2x%2B5+=+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=%282%29%5E2-4%2A-5%2A5=104.

Discriminant d=104 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+104+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+104+%29%29%2F2%5C-5+=+-0.819803902718557
x%5B2%5D+=+%28-%282%29-sqrt%28+104+%29%29%2F2%5C-5+=+1.21980390271856

Quadratic expression -5x%5E2%2B2x%2B5 can be factored:
-5x%5E2%2B2x%2B5+=+%28x--0.819803902718557%29%2A%28x-1.21980390271856%29
Again, the answer is: -0.819803902718557, 1.21980390271856. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-5%2Ax%5E2%2B2%2Ax%2B5+%29

Ignore the value >1
sin(y) =~-0.819804
y =~ -55 degs
or 305 degs, 235 degs
In radians, = 5.323, 4.102