# Thread: Trignometric identities

1. Can anybody please help me solve this?

4cot² - 6 cosec x = -6

2. Originally Posted by simongreen93
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

$\displaystyle 4\dfrac{\cos^2(x)}{\sin^2(x)} - \dfrac{6}{\sin(x)} = -6$

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?

3. Originally Posted by simongreen93
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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