MHB Equations that result in quadratics in some trigonometric function

[Simon green]

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

Solve the following equation for angles between 0˚ and 360˚ to 2 decimal places
4cos²θ + 5sinθ = 3
4 cot² - 6 Cosec x = -6

 

[topsquark]

Re: hyperbolic functions

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

Solve the following equation for angles between 0˚ and 360˚ to 2 decimal places
4cos²θ + 5sinθ = 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 [math]cos^2( \theta ) = 1 - sin^2( \theta )[/math], so the equation becomes:
[math]4 - 4~sin^2( \theta ) + 5~sin( \theta ) = 3[/math]

This is now a quadratic in [math]sin( \theta )[/math]

For the second start with [math]cot ( \theta ) = \frac{ cos( \theta )}{sin( \theta )}[/math] and [math]cosec( \theta ) = \frac{1}{sin( \theta )}[/math]

-Dan

 

[Prove It]

Re: hyperbolic functions

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

Solve the following equation for angles between 0˚ and 360˚ to 2 decimal places
4cos²θ + 5sinθ = 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 $\displaystyle \begin{align*} \cos^2{(\theta )} \end{align*}$.