Solve for x: 1 + sin2x = tan^2 x

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You should remember that `tan x = sin x/cos x` , hence, `tan^2 x = (sin x/cos x)^2`  => `tan^2 x = ((1-cos 2x)/2)/((1 + cos 2x)/2) = (1-cos 2x)/(1 + cos 2x)`

Hence, substituting  `(1 - cos 2x)/(1 + cos 2x)`  for `tan^2 x`  yields:

`1 + sin 2x =(1 - cos 2x)/(1 + cos 2x) ` `(1 + cos 2x)(1 + sin 2x) = 1 - cos 2x`

`1 + sin 2x + cos 2x + sin 2x cos 2x = 1 - cos 2x`

`sin2x + 2cos 2x + sin2x*cos 2x = 0`

`sin2x*(1 + cos 2x) + 2cos 2x = 0`

`sin2x*(1 + 2cos^2 x - 1) + 2cos 2x = ` 0

`4sin x*cos^3 x + 4cos^2x - 2 = 0`

`2sin x*cos^3 x + 2cos^2 x - 1 = 0`

`cos^2 x*(sin x*cos x + 1) = 1/2`

`cos^2 x = 1/2 => cosx = +-sqrt2/2=> x = +-pi/4 + 2npi`   `sin x*cos x + 1 = 1 => sin x*cos x = 0`

`sin x = 0 => x = npi`

`cos x = 0, x = +-pi/2 + 2npi`

Hence, evaluating the general solutions to the given equation, yields `x = +-pi/4 + 2npi ;x = +-pi/2 + 2npi ;x = npi.`  

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