How do I solve tan(x^2-1)=0?

Answer 1

y = tan(x^2 - 1) is not periodic, so my first answer will only refer to the interval -pi / 2 < x < pi / 2.

The tangent function equals zero at 0 radians.

tan(x^2 - 1) = 0
x^2 - 1 = 0
( x + 1)(x - 1) = 0
x + 1 = 0 or x - 1 = 0
x = 1 or x = -1

Answer: tan(x^2 - 1) = 0 at x = -1 and x = 1 in the interval -pi / 2 < x < pi / 2.

y = tan x is periodic and is equal to zero at multiples of pi.

Let k be an integer.

If tan(x^2 - 1) = 0, then x^2 - 1 = (pi)k.

x^2 = (pi)k + 1

x = sqrt[ (pi)k + 1 ] or x = -sqrt[ (pi)k + 1 ]

Since the square root function is not defined for negative values,
we let k >or = 0 so that (pi)k + 1 > 0.

Answer: For all real numbers x, tan(x^2 - 1) = 0 for x = sqrt[ (pi)k + 1 ] and x = sqrt[ (pi)k + 1 ] where integer k >or= 0.

Answer 2

TAN 0 radians = 0, so x^2-1=0, x=+1 and -1 radians.

Also additional solutions because of the repeating nature of TAN.

Answer 3

The tangent is 0 when the angle is or pi
Thus x^2-1 = 0 or pi
Thus x^2 = 1 --> x = 1 or -1 ,and
x^2-1 = pi --> x = +/- sqrt(1+pi)

Answer 4

We know that tan x = 0 when x = 0, pi, ..., n*pi.
Thus, take n*pi = (x^2) -1
so n * pi + 1 = x^2
sqrt(n*pi + 1) = x for integral n
Just make sure n * pi + 1 is positive, which means n=>0.

Answer 5

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

x^2-1 = atan(0) = n*pi n=...-2,-1,0,1,2...

x = sqrt(n*pi+1) and
x = -sqrt(n*pi+1)

if you are only concerned about real solutions...
then the sqrt function requires n*pi+1>0 or n>-1/pi~-1/3 but since n are always whole numbers, n=>-1 is sufficient. so your zeros are

x = sqrt(n*pi+1) and
x = -sqrt(n*pi+1)

for n=>-1

Answer 6

tan(x^2-1)=0
tan(x^2-1)=tan0
x^2-1=0
x^2=1
x=1 or x=-1

Answer 7

x² - 1 = 0
x² = 1
x = \/1
x = 1 or -1
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