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

Question 810405: Find all solutions t in the interval [0 , 2pi] to the equation sin(2t) = square root of 2 cos (t)
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find all solutions t in the interval [0 , 2pi] to the equation sin(2t) = square root of 2 cos (t)
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sin(2t)=√2cos(t)
2sin(t)cos(t)=√2cos(t)
square both sides
4sin^2(t)cos^2(t)=2cos^2(t)
divide by 2
2sin^2(t)cos^2(t)=cos^2(t)
divide by cos^2(t)
2sin^2(t)=1
sin^2(t)=1/2
sin(t)=�√(1/2)=�√2/2
t=π/4, 3π/4, 5π/4, 7π/4
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