- #1

AN630078

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- Homework Statement
- Hello, I have been practising solving trigonometric equations which are admittedly an area of mathematics which has caused me a great deal of confusion. In a textbook I came across the question below and quite liked the employment of the factor theorem in order to find solutions to the polynomial which is used to solve the trigonometric equation in the latter part of the question. However, I am still unsteady regarding trigonometric equations and wondered if anyone may be able to look over my solutions and offer any possible improvements or advice?

1. Use the factor theorem to factorise the expression: 3x^3-4x^2-5x+2

2. Hence solve the equation: 3cosec^3θ-4cosec^2θ-5cosecθ+2=0

- Relevant Equations
- 3x^3-4x^2-5x+2

3cosec^3θ-4cosec^2θ-5cosecθ+2=0

1. The factor theorem states that (x-a) is a factor of f(x) if f(a)=0

Therefore, suppose (x+1) is a factor:

f(-1)=3(-1)^3-4(-1)^2-5(-1)+2

f(-1)=0

So, (x+1) is a factor.

3x^3-4x^2-5x+2=(x+1)(3x^2+...)

Expand the RHS = 3x^3+3x^2

Leaving a remainder of -7x^2-5x+2

3x^3-4x^2-5x+2=(x+1)(3x^2-7x+...)

Expand the RHS = 3x^3+3x^2-7x^2-7x=3x^3-4x^2-7x

Leaving a remainder of 2x+2

3x^3-4x^2-5x+2=(x+1)(3x^2-7x+2)

Expand the RHS = 3x^3+3x^2-7x^2-7x+2x+2=3x^3-4x^2-5x+2

Factor (3x^2-7x+2) = (3x-1)(x-2)

The solutions are (x+1)(3x-1)(x-2)

x=-1, x=1/3,x=2

2. 3cosec^3θ-4cosec^2θ-5cosecθ+2=0

Let u=cosecθ

Therefore, 3u^3-4u^2-5u+2=0

Since we have found x=-1, x=1/3,x=2, thus u=1, u=1/3,u=2

When cosecθ=-1

Use the identity cosecθ=1/sinθ

Therefore, -1=1/sinθ

-sinθ=1

0=1+sinθ

sinθ=-1

θ=3π/2

cosecθ=1/3

1/3=1/sinθ

Apply cross multiplication;

a/b=c/d a*d=b*c

1*sinθ=3*1

sinθ=3

Which is a non-real solution since θ cannot be greater than 1 for real solutions.

cosecθ=2

Use the identity cosecθ=1/sinθ

Therefore, 2=1/sinθ

2sinθ=1

sinθ=1/2

θ=π/6, 5π/6

Thus, all the solutions are θ=3π/2+2πn, θ=π/6+2πn, θ=5π/6+2πn

Therefore, suppose (x+1) is a factor:

f(-1)=3(-1)^3-4(-1)^2-5(-1)+2

f(-1)=0

So, (x+1) is a factor.

3x^3-4x^2-5x+2=(x+1)(3x^2+...)

Expand the RHS = 3x^3+3x^2

Leaving a remainder of -7x^2-5x+2

3x^3-4x^2-5x+2=(x+1)(3x^2-7x+...)

Expand the RHS = 3x^3+3x^2-7x^2-7x=3x^3-4x^2-7x

Leaving a remainder of 2x+2

3x^3-4x^2-5x+2=(x+1)(3x^2-7x+2)

Expand the RHS = 3x^3+3x^2-7x^2-7x+2x+2=3x^3-4x^2-5x+2

Factor (3x^2-7x+2) = (3x-1)(x-2)

The solutions are (x+1)(3x-1)(x-2)

x=-1, x=1/3,x=2

2. 3cosec^3θ-4cosec^2θ-5cosecθ+2=0

Let u=cosecθ

Therefore, 3u^3-4u^2-5u+2=0

Since we have found x=-1, x=1/3,x=2, thus u=1, u=1/3,u=2

When cosecθ=-1

Use the identity cosecθ=1/sinθ

Therefore, -1=1/sinθ

-sinθ=1

0=1+sinθ

sinθ=-1

θ=3π/2

cosecθ=1/3

1/3=1/sinθ

Apply cross multiplication;

a/b=c/d a*d=b*c

1*sinθ=3*1

sinθ=3

Which is a non-real solution since θ cannot be greater than 1 for real solutions.

cosecθ=2

Use the identity cosecθ=1/sinθ

Therefore, 2=1/sinθ

2sinθ=1

sinθ=1/2

θ=π/6, 5π/6

Thus, all the solutions are θ=3π/2+2πn, θ=π/6+2πn, θ=5π/6+2πn