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I need help trying to sole tan^2 x =1 where x is more than or equal to 0 but x is less than or equal to pi

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3 Answers

Hi Brittany, 

The only numbers that square to one are 1 and -1.  

1 = 1 x 1

1 = (-1) x (-1)

So since (tan x) squares to one, (tan x) must be 1 or -1.  There are two values of x in the interval [0, pi] that make it happen.

tan (pi/4) = 1

tan (3*pi/4) = -1

Comments

    tan2(x) = 1 ,    0 ≤ x ≤ Π

tan2(x) = 1     ==>     (tan(x))2 = 1

√(tan(x))2 = √1     ==>     tan(x) = ± 1

Note the following:     tan(x) = sin(x)/cos(x)

So,   tan(x) = ± 1     ==>     sin(x)/cos(x) = ± 1

Multiply both sides of the equation by cos(x):

      (cos(x))·(sin(x)/cos(x)) = (cos(x))·(± 1)

        sin(x) = ± cos(x)

Looking at the top half of a unit circle (where x is between 0 and Π)...

...find the coordinates where sin(x) = cos(x) and sin(x) = -cos(x)

You will see that the coordinates that match are (√2/2, √2/2), which is located at x = ∏/4, and (-√2/2, √2/2), which is located at x = 3Π/4.

Thus,     x = Π/4    and     x = 3Π/4

Comments

Excellent explanation!