Trig help please!?

Question

1. (tan θ − 4)(25 sin2 θ − 1) = 0 

2. 3 tan3θ = tan θ 

Answer 1

1) The trigonometric equation is (tan θ − 4)(25 sin2 θ − 1) = 0.

Split the equations.

tan θ − 4 = 0 and 25 sin2 θ − 1 = 0

Take tan θ − 4 = 0.

Add 4 to each side.

tan θ − 4 + 4  = 0 + 4

tan θ = 4

Take the inverse tangent of both sides.

θ = tan^-1(4) + nπ for n Є Z.

Take 25 sin2θ − 1 = 0.

Add 1 to each side.

25 sin2θ − 1 + 1 = 0 + 1

25 sin2θ = 1

Divide each side by 25.

(25 sin2θ) / 25 = 1/25

sin2θ = 1/25

Take the invere sine of both sides.

2θ = 2πn ± sin^-1(1/25) for n Є Z.

2θ = 2πn₁ + sin^-1(1/25) n₁ Є Z or 2θ = 2πn₂ + π - sin^-1(1/25) n₂ Є Z.

θ = (1/2) {2πn₁ + sin^-1(1/25)} n₁ Є Z or θ = (1/2) {2πn₂ + π - sin^-1(1/25)} n₂ Є Z.

θ = πn₁ + (1/2)sin^-1(1/25) n₁ Є Z or θ = πn₂ + π/2 - (1/2)sin^-1(1/25) n₂ Є Z.

The solutions are θ = tan^-1(4) + nπ for n Є Z or θ = πn₁ + (1/2)sin^-1(1/25) n₁ Є Z or θ = πn₂ + π/2 - (1/2)sin^-1(1/25) n₂ Є Z.

 

Answer 2

2) The equation is 3 tan3θ = tan θ.

Formula for  .

Take Least Common Multiple.

Divide each side by 8.

Take inverse tangent of both sides.

The solution is .