Factor Theorem and Trigonometric Equations Help

In summary, the factor theorem states that (x-a) is a factor of f(x) if f(a)=0. Using this, we can determine that (x+1) is a factor of 3x^3-4x^2-5x+2. By expanding the right-hand side, we find that the remainder is 2x+2. Factoring (3x^2-7x+2), we get (3x-1)(x-2), leading to solutions of x=-1, x=1/3, and x=2. Similarly, for the equation 3cosec^3θ-4cosec^2θ-5cosecθ+2
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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
 
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  • #2
Looks good. Maybe a bit too detailed in parts, but well readable. As you end up with the sine function in each step, you could as well have multiplied the original equation by ##\sin^3 \theta## first, and work with the sine function alone. Also ##u=x## so one variable would have been sufficient.
 

Related to Factor Theorem and Trigonometric Equations Help

What is the Factor Theorem?

The Factor Theorem is a mathematical theorem that states that if a polynomial function has a root x=a, then (x-a) is a factor of the polynomial. This theorem is useful in finding the roots of polynomial equations and factoring them into simpler forms.

How do I use the Factor Theorem to solve equations?

To use the Factor Theorem to solve equations, you first need to identify the roots of the polynomial function. Then, you can use the roots to factor the polynomial and solve for the remaining variables.

What are trigonometric equations?

Trigonometric equations are equations that involve trigonometric functions such as sine, cosine, and tangent. These equations can be solved using various trigonometric identities and properties.

How do I solve trigonometric equations?

To solve trigonometric equations, you can use the unit circle, trigonometric identities, and solving techniques such as factoring and substitution. It is important to remember to check for extraneous solutions when solving trigonometric equations.

What is the difference between solving trigonometric equations and solving regular equations?

Solving trigonometric equations involves using trigonometric functions and identities, while solving regular equations typically involves algebraic operations such as addition, subtraction, multiplication, and division. Additionally, trigonometric equations may have multiple solutions due to the periodic nature of trigonometric functions.

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