Question 476614: Find the solution of the equation for [0,2pi]
2(tan(theta))-(sec^2(theta))=0
Here is what I've done so far:
2tan(x) = sec^2(x)---> I set them equal because of some rule? and replaced theta w/x to make it easier for me to understand.
2(sin(x)/cos(x)) = 1/cos(x)
2 = 1/cos(x) * (cos(x)/sin(x))
2(cos(x)*sin(x)) = 1
(cos(x)*sin(x)) = 1/2
x = ?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the solution of the equation for [0,2pi]
2(tan(theta))-(sec^2(theta))=0
Here is what I've done so far:
2tan(x) = sec^2(x)---> I set them equal because of some rule? and replaced theta w/x to make it easier for me to understand.
2(sin(x)/cos(x)) = 1/cos(x) **** What happened to the square on the RHS?
2 = 1/cos(x) * (cos(x)/sin(x))
2(cos(x)*sin(x)) = 1
(cos(x)*sin(x)) = 1/2
x = ?
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2(tan(theta))-(sec^2(theta))=0
2sin(x)/cos(x) - 1/cos^2(x) = 0
Multiply by cos^2
2sin(x)cos(x) - 1 = 0
sin(2x) - 1 = 0
sin(2x) = 1
2x = pi/2, 5pi/2
x = pi/4, 5pi/4
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You would have gotten it, but you lost the square of the RHS term.
Then you have to recognize the double angle on the LHS,
sin(2x) = 2sin(x)cos(x)
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