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

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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).
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Can anybody please help me solve this?

4cot² - 6 cosec x = -6
 
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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.
 
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