# Can someone help me with a Trig problem?

+1 vote

The problem reads: solve the equation in the given domain: 0<x<2pi

tan2x = -1

I already have the answer sheet, I just can't figure out how to do it. Could someone show me the process?

tan 2x = -1

Trigonometry table in tan 135° = -1

So, tan 2x = tan 135°

Cancel common term.

2x = 135°

Divide each side by 2.

x = 67.5 = 3π / 8.

The general solution is x = 1/2(nπ - π/4) where n Є Z

and the solutions are 3π/8, 7π/8, 11π/8, 15π/8.

The equation is tan2x  = -1.

tan2x  = tan(-π/4)

Apply inverse tangent to each side.

If tanθ = tanα, then θ = nπ + α.

2x  = nπ + (-π/4) for n Є Z.

2x  = nπ - π/4 for n Є Z.

x  = 1/2 (nπ - π/4) for n Є Z.

The interval is 0 < x < 2π, means x  lies between 0 and 2π.

Substitute 1 for n, x  = 1/2 (1*π - π/4) = 1/2 (3π/4) = 3π/8 < 2π

Substitute 2 for n, x  = 1/2 (2*π - π/4) = 1/2 (7π/4) = 7π/8 < 2π

Substitute 3 for n, x  = 1/2 (3*π - π/4) = 1/2 (11π/4) = 11π/8 < 2π

Substitute 4 for n, x  = 1/2 (4*π - π/4) = 1/2 (15π/4) = 15π/8 < 2π

Substitute 5 for n, x  = 1/2 (5*π - π/4) = 1/2 (19π/4) = 19π/8 > 2π

Therefore the solutions are 3π/8, 7π/8, 11π/8 and 15π/8.

The trigonometric equation is tan(2x) = - 1 and the domain is : 0 < x < 2π.

Let 2x = t, the function tan(t) has a period of π, first find all solutions in the interval [ 0, π ).

tan(t) = - 1

t = tan- 1( - 1 ).

The function tan(x) is negative in second and fourth quadrant.

- 1 = - tan (π/4) = tan (π/2 + π/4) = tan (2π + π)/4 = tan (3π/4).

In fourth Quadrant, 3π/2x ≤ 2π, there is no solution exist because  tan(x) has period π.

So the solutions are

t = tan- 1(- 1) -----> t = tan-1[tan (3π/4)] -------> t = 3π/4.

The solutions x = 3π/8 in the interval [ 0, 2π).

Finally, add multiples of π to each of these solutions to get the general form t = 3π/4 + nπ = π(3/4 + n), where n is an integer.

2x = π(3/4 + n) -----> x = π(3/8 + n/2)

The solutions outside the interval [ 0, 2π ) are

If n = - 1 then x = π[ 3/8 + (- 1)/2 ] = - π/8 < 0.

If n = 4 then x = π[ 3/8 + (4)/2 ] = 19π/8 > 0.

The solutions in the interval [ 0, 2π ) are

If n = 0 then x = π[ 3/8 + (0)/2 ] = 3π/8.

If n = 1 then x = π[ 3/8 + (1)/2 ] = 7π/8.

If n = 2 then x = π[ 3/8 + (2)/2 ] = 11π/8.

If n = 3 then x = π[ 3/8 + (3)/2 ] = 15π/8.

The solutions in the interval [ 0, 2π ) are 3π/8, 7π/8, 11π/8 and 15π/8.