Can anybody please help me solve this?
4cotē - 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?
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.