SOLUTION: Solver the given trig equation exactly on 0<=theta<2pi
question is: 2cos(theta/2)= -sqrt2
So far I divided the 2 on the left and the right side of the equation leaving me wit
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-> SOLUTION: Solver the given trig equation exactly on 0<=theta<2pi
question is: 2cos(theta/2)= -sqrt2
So far I divided the 2 on the left and the right side of the equation leaving me wit
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Question 980560: Solver the given trig equation exactly on 0<=theta<2pi
question is: 2cos(theta/2)= -sqrt2
So far I divided the 2 on the left and the right side of the equation leaving me with, Cos(theta/2)= -sqrt2/2
Now, I'm lost...
(this is a pre-cal question. The topic list is blank for me. Sorry)
You can put this solution on YOUR website! Solver the given trig equation exactly on 0<=theta<2pi
question is: 2cos(theta/2)= -sqrt2
So far I divided the 2 on the left and the right side of the equation leaving me with, Cos(theta/2)= -sqrt2/2
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Use the double angle formula:
cos(2t) = 2cos^2(t) - 1
= 2*(1/2) - 1
cos(theta) = 0
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theta = pi/2, 3pi/2
