1. I have absolutely no idea how to tackle either of these questions

Solve the following equation for angles between 0Ëš and 360Ëš to 2 decimal places
4cos²Î¸ + 5sinÎ¸ = 3
4 cot² - 6 Cosec x = -6

2. Originally Posted by simongreen93
I have absolutely no idea how to tackle either of these questions

Solve the following equation for angles between 0Ëš and 360Ëš to 2 decimal places
4cos²Î¸ + 5sinÎ¸ = 3
4 cot² - 6 Cosec x = -6
This has nothing to do with the hyperbolic functions!?

To start you off:

For the first, recall that $\displaystyle cos^2( \theta ) = 1 - sin^2( \theta )$, so the equation becomes:
$\displaystyle 4 - 4~sin^2( \theta ) + 5~sin( \theta ) = 3$

This is now a quadratic in $\displaystyle sin( \theta )$

For the second start with $\displaystyle cot ( \theta ) = \frac{ cos( \theta )}{sin( \theta )}$ and $\displaystyle cosec( \theta ) = \frac{1}{sin( \theta )}$

-Dan

3. Originally Posted by simongreen93
I have absolutely no idea how to tackle either of these questions

Solve the following equation for angles between 0Ëš and 360Ëš to 2 decimal places
4cos²Î¸ + 5sinÎ¸ = 3
4 cot² - 6 Cosec x = -6
First of all, these are trigonometric functions, not hyperbolic.

Second, both are solved in the exact same way. Use the Pythagorean Identity (or some variation of it) to rewrite your equation all in terms of the same trigonometric function. Then solve the resulting quadratic equation.

So in the first for example, you will substitute \displaystyle \begin{align*} 1 - \sin^2{(\theta )} \end{align*} for \displaystyle \begin{align*} \cos^2{(\theta )} \end{align*}.