Brittany A.

asked • 11/15/12# 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

this is what it looks like tan^2 x=1 where 0 ≤ x ≤ π

## 4 Answers By Expert Tutors

Tamara J. answered • 11/15/12

Math Tutoring - Algebra and Calculus (all levels)

tan^{2}(x) = 1 , 0 ≤ x ≤ Π

tan^{2}(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

Elena B.

Excellent explanation!

11/18/12

Robert J. answered • 11/15/12

Certified High School AP Calculus and Physics Teacher

Solve for tan x first, tan x = ±1

Therefore, x = pi/4, 3pi/4 on [0, pi]

Osman A. answered • 30d

Professor of Engineering Mathematics – Trigonometry and Geometry

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. this is what it looks like tan^2 x=1 where 0 ≤ x ≤ π

__Detailed Solution:__

tan^{2} x = 1 0 ≤ x ≤ π

tan x = ±1

tan x = 1 (Quadrant 1 only in 0 ≤ x ≤ π)

**x = tan**^{-1}** 1 = π/4**

tan x = –1 (Quadrant 2 only in 0 ≤ x ≤ π)

**x = tan**^{-1}** –1 = π – π/4 = 3π/4**

__x = (π/4, 3π/4)__

Raymond B. answered • 04/10/21

Math, microeconomics or criminal justice

tan^2(x) = 1

take the square root of both sides

tanx = + or -1 at this point you could recognize it's a 1-1- sqr2 right triangle that's 45 degrees Or you could plug in 1 into a calculator with an inverse trig function, and it would read 45 degrees or pi/4, plug in -1 and the inverse tangent will probably say -45 degrees. Add 180 to -45 to get 135 degrees. You want quadrant I and II, not IV where -45 degrees is.

answer is 45 and 135 degrees or pi/4 and 3pi/4 radians

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Brittany A.

thanks

11/15/12