2 * tan(x)^2 - 1 = 0, solve for x between 0->2pi2 * tan(x)^2 - 1 = 0 2 * tan(x)^2 = 1 tan(x)^2 = 1/2 tan(x) = +/- *sqrt(2) / 2 Lost from here!

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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2tan^2 x -1 = 0

==> 2tan^2 x = 1

==> tan^2 x = 1/2

Now we know that tanx = sinx/cosx

==> tan^2 x = sin^2 x/ cos^2 x

==> (sin^2 x) / (cos^2 x) = 1/2

Now we know that cos^2 x = 1- sin^2 x

=> sin^2 x / (1-sin^2 x) = 1/2

==> 2sin^2 x = 1- sin^2 x

==> 3sin^2 x = 1

==> sin^2 x = 1/3

==> sinx = +-sqrt(1/3) =

==> sinx =+- 0.5774

==> x1 = +-35.26 degrees.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve 2 * tan(x)^2 - 1 = 0 for x in [0, 2*pi]

2 * tan(x)^2 - 1 = 0

=> 2 * tan(x)^2 = 1

=> tan(x)^2 = 1/2

=> tan x = 1/sqrt 2

=> x = arc tan (1/sqrt 2)

=> x = 35.2643 and x = 215.2634

The required solutions within [0, 2pi] are x = 35.2643 degrees and x = 215.2643 degrees.

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