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)   (Show Source): You can put this solution on YOUR website!
2 tan y + 5 cosy y = 0 for 0 < y < 6
---------------------
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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=104 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -0.819803902718557, 1.21980390271856. Here's your graph:

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

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