SOLUTION: Find all solutions to the equation on the interval [0,2pie) cos(2theta)-tan(theta)=1
Here is my work: *=theta
cos(2*)-tan(*)-1=0
1-2sin^2(*)-sin(*)/cos(*)-1=0
cos(*)-2sin^2(*
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-> SOLUTION: Find all solutions to the equation on the interval [0,2pie) cos(2theta)-tan(theta)=1
Here is my work: *=theta
cos(2*)-tan(*)-1=0
1-2sin^2(*)-sin(*)/cos(*)-1=0
cos(*)-2sin^2(*
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Question 78336: Find all solutions to the equation on the interval [0,2pie) cos(2theta)-tan(theta)=1
Here is my work: *=theta
cos(2*)-tan(*)-1=0
1-2sin^2(*)-sin(*)/cos(*)-1=0
cos(*)-2sin^2(*)cos(*)-sin(*)-cos(*)=0
-2sin^2(*)cos(*)-sin(*)=0
-Sin(*)[2sin(*)cos(*)-1]=0
-sin(*)=0 2sin(*)cos(*)=0
-(*)=-pie/2,-3pie/2 sin(*)=0 cos(*)=0
(*)=pie/2,3pie/2 (*)=pie, 2pie
Did I do this correctly? Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! You're right so far, but lets look at this step "-2sin^2(*)cos(*)-sin(*)=0"
Factor out a Set the first term equal to zero
Take the arcsine of both sides
So for this solution you should have
(if we let n=0), (if we let n=1)
Now let equal zero Replace with Add 1 to both sides
Take the arc sine of both sides
Divide both sides by -2
So just looking in the interval [0,) (excluding ) we get the solutions
Let n=1
Let n=2
So we only have 4 solutions in the interval [0,):
, , ,