MHB Equations that result in quadratics in some trigonometric function

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The discussion revolves around solving two trigonometric equations: 4cos²θ + 5sinθ = 3 and 4cot²x - 6cosecx = -6. Participants clarify that these are trigonometric, not hyperbolic functions, and suggest using the Pythagorean Identity to rewrite the equations in terms of a single trigonometric function. The first equation can be transformed into a quadratic in sin(θ) by substituting cos²(θ) with 1 - sin²(θ). The approach for both equations involves solving the resulting quadratic equations to find angles between 0˚ and 360˚.
Simon green
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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
 
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Re: hyperbolic functions

simongreen93 said:
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
 
Re: hyperbolic functions

simongreen93 said:
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 $\displaystyle \begin{align*} \cos^2{(\theta )} \end{align*}$.
 
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