SOLUTION: Find all values of t in the interval [0, 2π] satisfying the given equation: 6 sin 2t + 3 tan 2t = 0. And 6 is under a square root! How to get the answer? Thank you!

Algebra ->  Trigonometry-basics -> SOLUTION: Find all values of t in the interval [0, 2π] satisfying the given equation: 6 sin 2t + 3 tan 2t = 0. And 6 is under a square root! How to get the answer? Thank you!      Log On


   

Question 805610: Find all values of t in the interval [0, 2π] satisfying the given equation: 6 sin 2t + 3 tan 2t = 0. And 6 is under a square root! How to get the answer? Thank you!
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find all values of t in the interval [0, 2π] satisfying the given equation:
6 sin 2t + 3 tan 2t = 0. And 6 is under a square root
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sqrt%286%29%2Asin%282t%29+%2B+3tan%282t%29+=+0
Use x for 2t
sqrt%286%29%2Asin%28x%29+%2B+3sin%28x%29%2Fcos%28x%29+=+0
sin%28x%29%2A%28sqrt%286%29+%2B+3%2Fcos%28x%29%29+=+0
sin(x) = 0
x = 2t = 0, pi, 2pi
t = 0, pi/2, pi
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sqrt%286%29+%2B+3%2Fcos%28x%29+=+0
sqrt%286%29%2Acos%28x%29+%2B+3+=+0
cos%28x%29+=+-3%2Fsqrt%286%29
|cos(x)| > 1, no real number solution