MHB How can the quadratic equation be used to solve a trigonometric identity?

  • Thread starter Thread starter Simon green
  • Start date Start date
  • Tags Tags
    identities
AI Thread Summary
The discussion focuses on solving the equation 4cot² - 6 cosec x = -6 by converting it into a quadratic form. The transformation involves rewriting the equation in terms of sine, leading to the expression 4(csc²(x) - 1) - 6csc(x) = -6. By applying the identity 1 + cot²(x) = csc²(x), the equation simplifies to a standard quadratic form: 2csc²(x) - 3csc(x) + 1 = 0. Participants suggest clearing the denominator and working with a single trigonometric function for easier manipulation. The final step involves solving the quadratic equation for csc(x).
Simon green
Messages
10
Reaction score
0
Can anybody please help me solve this?

4cot² - 6 cosec x = -6
 
Mathematics news on Phys.org
simongreen93 said:
Can anybody please help me solve this?

4cot² - 6 cosec x = -6

Good evening, what thoughts have you had to help solve this problem?I prefer to write everything in terms of sin and cos

[math]4\dfrac{\cos^2(x)}{\sin^2(x)} - \dfrac{6}{\sin(x)} = -6[/math]

Some thoughts:

  • Clear the denominator by multiplying by the LCD of the terms above
  • Work only with one trig function - I would recommend sine as you have more of them - do you know of an identity to change your cos to sin?
 
simongreen93 said:
Can anybody please help me solve this?

4cot² - 6 cosec x = -6

Since $\displaystyle \begin{align*} 1 + \cot^2{(x)} \equiv \csc^2{(x)} \end{align*}$ that means

$\displaystyle \begin{align*} 4\left[ \csc^2{(x)} - 1 \right] - 6\csc{(x)} &= -6 \\ 4\csc^2{(x)} - 4 - 6\csc{(x)} &= -6 \\ 4\csc^2{(x)} - 6\csc{(x)} + 2 &= 0 \\ 2\csc^2{(x)} - 3\csc{(x)} + 1 &= 0 \end{align*}$

Now solve the resulting quadratic.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top