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What values of x in `[0, 2*pi)` satisfy the following equation: `sec x+tanx=1`
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The equation sec x + tan x = 1 has to be solved for x in `[0, 2*pi)`
sec x + tan x = 1
Take the square of both the sides. (Note: This will add some roots that are roots of the equation sec x + tan x = -1 which have to be eliminated later)
`sec^2 x + tan^2 x + 2*(sin x/cos x)*(1/cos x)= 1`
=> `(1/cos x)^2 +(sin x/cos x)^2 + 2*(sin x/cos x)*(1/cos x)= 1`
=> `(1 + sin^2x + 2*sin x)/cos^2x = 1`
=> `(1 + sin^2x + 2*sin x) = cos^2 x`
=> `(1 + sin^2x + 2*sin x) = 1 - sin^2x `
=> `2*sin^2x + 2*sin x = 0`
=> sin x = 0 and sin x = -1
=> x = 0, x = `pi` , x = `(3*pi)/2`
But tan x is infinite for `x = (3*pi)/2` and the value of `sec pi + tan pi = -1`
The solution of the equation is x = 0.
Posted by justaguide on May 5, 2013 at 6:41 PM (Answer #1)
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