# Help me with my trig homework?

Solve

1. csc^2(X/2) = 2secX,   -Pi<(or equal to) X < Pi

2. Tan^2X-|secX|=1, X belongs to [0,2pi)

1.

csc2(x / 2) = 2secx , [-π, π]

1 / sin2(x / 2) = 2secx

1 / sin(x /2) = 2sin(x / 2) / cosx

cosx = 2sin(x / 2) sin(x / 2)

Multiply each side by cos(x / 2)

cos(x / 2)cosx = sin(x / 2)sinx

cosx cos(x / 2) = sinx sin(x / 2)

cosx cos(x / 2) - sinx sin(x / 2) = 0

Recall : cos(A + B) = cosAcosB - sinAsinB and cos(π / 2) = 0 ,cos(-π / 2) = 0

cos(x + (x / 2)) = cos(π / 2)

x + x / 2 = π / 2 or x + x /2 = -π / 2

3x / 2 = π / 2 or 3x / 2 = -π / 2

3x = π or 3x = -π

x = π / 3 or x = -π / 3.

cos(3x/2) = cos(π/2).

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

3x/2 = 2nπ ± π/2

⇒ x = (4nπ/3) ± π/3.

If n = 0, x = (4*0*π/3) ± π/3 = ± π/3,

If n = 1, x = (4*1*π/3) ± π/3 = (4π/3) + π/3  and  (4π/3) - π/3 = 5π/3  and  π.

Therefore, the solutions of the given equation are x = - π/3 and x = π/3 in the interval [- π, π).

1.

tan2x -|secx| = 1

sec2x - 1 - |secx| = 1

sec2x  - |secx| = 2

sec2x  - |secx| - 2= 0

case 1:

sec2x - secx - 2 = 0

sec2x -2secx + secx - 2 = 0

secx(secx - 2) + 1(secx - 2) = 0

(secx -2)(secx + 1) = 0

secx - 2 = 0

secx = 2

cosx = 1 / 2

x = π/3 x € [0, 2π)

if secx + 1 = 0 then

secx = -1

cosx = -1

x =π

Case 2 :

sec2x  + secx - 2 = 0

sec2x + 2secx - secx  - 2 = 0

secx(secx + 2) -1(secx + 2) = 0

secx - 1 = 0

secx = 1

cosx = 1

x = 0  x belongs to[0, 2pi)

secx + 2 = 0

secx = -2

cosx = -1 / 2

x = 2π/ 3

The values are 0 , π/3 , 2π / 3 ,  pi .

Case 1 : (sec x - 2)(sec x + 1) = 0

• sec x = 2.

Using reciprocal identity : sec x = 1/cos x.

⇒ cos x = 1/2.

cos x = cos (π/3).

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒x = 2nπ ± (π/3).

If n = 0, x = 2(0)π + (π/3) and x = 2(0)π - (π/3) = π/3 and - π/3,

If n = 1, x = 2(1)π + (π/3) and x = 2(1)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

x = π/3 and x = 5π/3 in the interval [0, 2π).

• sec x = - 1

cos x = - 1.

cos (x) = cos(π)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒x = 2nπ ± π

If n = 0, x = 2(0)π + π and x = 2(0)π - π = π and - π,

If n = 1, x = 2(1)π + π and x = 2(1)π - π = 2π + π and 2π - π = and π.

x = π in the interval [0, 2π).

Case 2 : (sec x + 2)(sec x - 1) = 0.

• sec x = - 2.

⇒ cos x = - 1/2.

cos x = cos (2π/3).

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒x = 2nπ ± (2π/3).

If n = 0, x = 2(0)π + (2π/3) and x = 2(0)π - (2π/3) = 2π/3 and - 2π/3,

If n = 1, x = 2(1)π + (2π/3) and x = 2(1)π - (2π/3) = 2π + 2π/3 and 2π - 2π/3 = 8π/3 and 4π/3.

x = 2π/3 and x = 4π/3 in the interval [0, 2π).

• sec x = 1

cos x = 1.

cos (x) = cos(0)

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

⇒x = 2nπ ± 0

If n = 0, x = 2(0)π = 0,

x = 0 in the interval [0, 2π).

Therefore, the solutions of the given equation are x = 0, x = π/3, x = 2π/3, x = π, x = 4π/3, and x = 5π/3 in the interval [0, 2π).