SOLUTION: Find all values of t in the interval [0, 2𝜋] satisfying the given equation. (Enter your answers as a comma-separated list.) 2 ((square root)2)sin 2t − ((square root)6)tan 2t

Algebra ->  Trigonometry-basics -> SOLUTION: Find all values of t in the interval [0, 2𝜋] satisfying the given equation. (Enter your answers as a comma-separated list.) 2 ((square root)2)sin 2t − ((square root)6)tan 2t       Log On


   

Question 1170579: Find all values of t in the interval [0, 2𝜋] satisfying the given equation. (Enter your answers as a comma-separated list.)
2 ((square root)2)sin 2t − ((square root)6)tan 2t = 0
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
2sqrt%282%29%2Asin%282t%29+-+sqrt%286%29%2Atan%282t%29+=+0
let 2t+=+x
2sqrt%282%29%2Asin%28x%29+-+sqrt%286%29%2Atan%28x%29+=+0..............tan%28x%29=sin+%28x%29%2Fcos%28x%29
2sqrt%282%29%2Asin%28x%29+-+sqrt%286%29%28sin+%28x%29%2Fcos%28x%29%29+=+0............times cos(x)
2sqrt%282%29%2Asin%28x%29cos%28x%29+-+sqrt%286%29%2Asin%28x%29+=+0
%282sqrt%282%29%2Acos%28x%29+-+sqrt%286%29%29%2Asin+%28x%29+=+0

=> sin+%28x%29+=+0 =>sin+%282t%29+=+0
t in the interval [0, 2pi] : t=0, t=+pi%2F2, t=+pi , t=+3pi%2F2, t=2pi
or
2sqrt%282%29%2Acos%28x%29+-+sqrt%286%29=+0=>cos%28x%29+=+sqrt%286%29%2F2sqrt%282%29
=>cos%282t%29+=+sqrt%286%29%2F2sqrt%282%29
=>cos%282t%29+=+sqrt%283%29%2F2
2t=cos%5E-1%28+sqrt%283%29%2F2%29
2t=pi%2F6+
t=pi%2F12+

t in the interval [0,+2pi] : t=pi%2F12,t=11pi%2F12,t=13pi%2F12,t=23pi%2F12

combine solutions:
t=0, t=+pi%2F2, t=+pi , t=+3pi%2F2, t=2pi,t=pi%2F12,t=11pi%2F12,t=13pi%2F12,t=23pi%2F12