SOLUTION: Find all the solutions t in the interval (0, 2pi) to the equation sin (2t) = sqare root 2 cos (t)

Question 598457: Find all the solutions t in the interval (0, 2pi) to the equation
sin (2t) = sqare root 2 cos (t)
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find all the solutions t in the interval (0, 2pi) to the equation
sin (2t) = sqare root 2 cos (t)
2sintcost=√2cost
2sintcost-√2cost=0
cost(2sint-√2)=0
cost=0
t=π/2 and 3π/2
or
2sint-√2=0
sint=√2/2
t=π/4 and 3π/4 (in quadrants I and II where sin>0)
..
solutions: π/4, π/2, 3π/4 and 3π/2

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